course phy 201 ???~s??^????`???z????assignment #030
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11:01:59 introductory set 8. If we know the constant moment of inertia of a rotating object and the constant net torque on the object, then how do we determine the angle through which it will rotate, starting from rest, in a given time interval?
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RESPONSE --> Know constant moment of inertia ""I"" Know net torque How much will it rotate from rest in given time period? net tau = I * alpha can be rearranged to alpha = net tau / I so dividing net torque by moment of inertia will give us the angular acceleration. having this we can then use the formula: vfinal = v0 + alpha * dt
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11:02:50 ** tau stands for torque and I stands for the moment of inertia. These quantities are analogous to force and mass. Just as F = m a, we have tau = I * alpha; i.e., torque = moment of inertia * angular acceleration. If we know the moment of inertia and the torque we can find the angular acceleration. If we multiply angular acceleration by time interval we get change in angular velocity. We add the change in angular velocity to the initial angular velocity to get the final angular velocity. In this case initial angular velocity is zero so final angular velocity is equal to the change in angular velocity. If we average initial velocity with final velocity then, if angular accel is constant, we get average angular velocity. In this case angular accel is constant and init vel is zero, so ave angular vel is half of final angular vel. When we multiply the average angular velocity by the time interval we get the angular displacement, i.e., the angle through which the object moves. **
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RESPONSE --> ok
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11:05:17 If we know the initial angular velocity of a rotating object, and if we know its angular velocity after a given time, then if we also know the net constant torque accelerating the object, how would we find its constant moment of inertia?
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RESPONSE --> moment of inertia = net torque / angular accel by taking the change in angular velocity / time we get angular accel. we are given the net torque, so we can plug these into the formula above to solve for moment of inertia.
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11:05:42 ** From init and final angular vel you find change in angular vel (`d`omega = `omegaf - `omega0). You can from this and the given time interval find Angular accel = change in angular vel / change in clock time. Then from the known torque and angular acceleration we find moment of intertia. tau = I * alpha so I = tau / alpha. **
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RESPONSE --> ok
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11:13:39 How do we find the moment of inertia of a concentric configuration of 3 uniform hoops, given the mass and radius of each?
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RESPONSE --> take the mass of each ring times the square of the radius from the point of rotation and total these three quantities.
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11:14:43 How do we find the moment of inertia a light beam to which are attached 3 masses, each of known mass and lying at a known distance from the axis of rotation?
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RESPONSE --> multiply each mass times the distance from the point of rotation and then add these three quantities. Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems.
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11:14:54 ** Moment of inertia of a mass r at distance r is m r^2. We would get a total of m1 r1^2 + m2 r2^2 + m3 r3^2. Note the similarity to the expression for the hoops. **
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RESPONSE --> ok
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11:17:19 Principles of Physics and General College Physics problem 8.4. Angular acceleration of blender blades slowing to rest from 6500 rmp in 3.0 seconds.
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RESPONSE --> blades drop from 6500 rpm to zero in 3sec. 6500 / 3 = - 2167 rpm/s
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11:18:12 The change in angular velocity from 6500 rpm to rest is -6500 rpm. This change occurs in 3.0 sec, so the average rate of change of angular velocity with respect to clock time is ave rate = change in angular velocity / change in clock time = -6500 rpm / (3.0 sec) = -2200 rpm / sec. This reasoning should be very clear from the definition of average rate of change. Symbolically the angular velocity changes from omega_0 = 6500 rpm to omega_f = 0, so the change in velocity is `dOmega = omega_f - omega_0 = 0 - 6500 rpm = -6500 rpm. This change occurs in time interval `dt = 3.0 sec. The average rate of change of angular velocity with respect to clock time is therefore ave rate = change in angular vel / change in clock time = `dOmega / `dt = (omega_f - omega_0) / `dt = (0 - 6500 rpm) / (3 sec) = -2200 rpm / sec. The unit rpm / sec is a perfectly valid unit for rate of change of angular velocity, however it is not the standard unit. The standard unit for angular velocity is the radian / second, and to put the answer into standard units we must express the change in angular velocity in radians / second. Since 1 revolution corresponds to an angular displacement of 2 pi radians, and since 60 seconds = 1 minute, it follows that 1 rpm = 1 revolution / minute = 2 pi radians / 60 second = pi/30 rad / sec. Thus our conversion factor between rpm and rad/sec is (pi/30 rad / sec) / (rpm) and our 2200 rpm / sec becomes angular acceleration = 2200 rpm / sec * (pi/30 rad / sec) / rpm = (2200 pi / 30) rad / sec^2 = 73 pi rad / sec^2, or about 210 rad / sec^2.
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RESPONSE --> ok
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11:23:58 Principles of Physics and General College Physics problem 8.16. Automobile engine slows from 4500 rpm to 1200 rpm in 2.5 sec. Assuming constant angular acceleration, what is the angular acceleration and how how many revolutions does the engine make in this time?
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RESPONSE --> initial speed = 4500 rpm final speed = 1200 rpm d speed = 3300 rpm dt = 2.5 sec angular accel? total number of revolutions? 3300 / 2.5 = - 1320 rpm/sec angular accel (4500 + 1200)/ 2 = 2850 rpm (171,000 rps) avg speed * 2.5 sec = 4.27 *10^5 revolutions
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11:31:06 The change in angular velocity is -3300 rpm, which occurs in 2.5 sec. So the angular acceleration is angular accel = rate of change of angular vel with respect to clock time = -3300 rpm / (2.5 sec) = 1300 rpm / sec. Converting to radians / sec this is about angular accel = -1300 rpm / sec ( pi / 30 rad/sec) / rpm = 43 pi rad/sec^2, approx.. Since angular acceleration is assumed constant, a graph of angular velocity vs. clock time will be linear so that the average angular velocity with be the average of the initial and final angular velocities: ave angular velocity = (4500 rpm + 1200 rpm) / 2 = 2750 rpm, so that the angular displacement is angular displacement = ave angular velocity * time interval = 2750 rpm * 2.5 sec = 6900 revolutions, approximately. In symbols, using the equations of uniformly accelerated motion, we could use the first equation `dTheta = (omega_0 + omega_f) / 2 * `dt = (4500 rpm + 1200 rpm) / 2 * (2.5 sec) = 6900 revolutions and the second equation omega_f = omega_0 + alpha * `dt, which is solved for alpha to get alpha = (omega_f - omega_0) / `dt = (4500 rpm - 1200 rpm) / (2.5 sec) = 1300 rpm / sec, which as before can be converted to about 43 pi rad/sec^2, or about 130 rad/sec^2. The angular displacement of 6900 revolutions can also be expressed in radians as 6900 rev = 6900 rev (2 pi rad / rev) = 13800 pi rad, or about 42,000 radians.
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RESPONSE --> I think there are a couple of errors here: first, (4500 + 1200) / 2 is 2850 rpm not 2750 next: I don't think we can multiply rpm's times dt in sec to get # of revolutions. Shouldn't it be 2850 rpm/60 = 47.5 rps * 2.5 sec = 118.75 total revolutions? I made an error in multiplying the rpm's times 60 instead of dividing when I converted them to rps.
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11:35:04 gen Problem 8.23: A 55 N force is applied to the side furthest from the hinges, on a door 74 cm wide. The force is applied at an angle of 45 degrees from the face of the door. Give your solution:
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RESPONSE --> the net tau = force * moment arm * sin theta (45deg) = 55N * .74m * .707 = 28.7 m*N
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11:36:05 ** ** If the force is exerted perpendicular to the face of the door, then the torque on the door is 55 N * .74 m = 40.7 m N. The rest of the given solution here is for a force applied at an angle of 60 degrees. You can easily adapt it to the question in the current edition, where the angle is 45 degrees:The torque on the door is 45 N * .84 m = 37.8 m N. If the force is at 60 deg to the face of the door then since the moment arm is along the fact of the door, the force component perpendicular to the moment arm is Fperp = 37.8 m N * sin(60 deg) = 32.7 N and the torque is torque = Fperp * moment arm = 32.7 N * .84 m = 27.5 m N. STUDENT COMMENT: Looks like I should have used the sin of the angle instead of the cosine. I was a little confused at which one to use. I had trouble visualizing the x and y coordinates in this situation.INSTRUCTOR RESPONSE: You can let either axis correspond to the plane of the door, but since the given angle is with the door and angles are measured from the x axis the natural choice would be to let the x axis be in the plane of the door. The force is therefore at 60 degrees to the x axis. We want the force component perpendicular to the door. The y direction is perpendicular to the door. So we use the sine of the 60 degree angle.
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RESPONSE --> ok
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11:45:40 gen problem 8.11 rpm of centrifuge if a particle 7 cm from the axis of rotation experiences 100,000 g's
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RESPONSE --> I was very confused by this one. Is the 100,000 g's the tangential accel? If yes, then setting the 100,000 g's = to r * (d angular velocity/dt) and rewriting it to solve for a tang should give me the answer but what is the dt?
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11:47:00 ** alpha = v^2 / r so v = `sqrt( alpha * r ) = `sqrt( 100,000 * 9.8 m/s^2 * .07 m) = `sqrt( 69,000 m^2 / s^2 ) = 260 m/s approx. Circumference of the circle is 2 `pi r = 2 `pi * .07 m = .43 m. 260 m/s / ( .43 m / rev) = 600 rev / sec. 600 rev / sec * ( 60 sec / min) = 36000 rev / min or 36000 rpm. All calculations are approximate. **
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RESPONSE --> So it was the centripetal accel that we wanted? not the tangential accel?
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11:56:55 gen problem 8.20 small wheel rad 2 cm in contact with 25 cm wheel, no slipping, small wheel accel at 7.2 rad/s^2. What is the angular acceleration of the larger wheel?
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RESPONSE --> The bigger wheel has a radius and, therefore, a cicumference which is 12.5 times as great as the small wheel. Therefore the large wheel's angular acceleration will be (7.2 rad/sec^2) / 12.5 = .576 rad /sec^2
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11:57:14 ** Since both wheels travel the same distances at the rim, angular displacements (which are equal to distance along the rim divided by radii) will be in inverse proportion to the radii. It follows that angular velocities and angular accelerations will also be in inverse proportion to radii. The angular acceleration of the second wheel will therefore be 2/25 that of the first, or 2/25 * 7.2 rad/s^2 = .58 rad/s^2 approx.. **
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RESPONSE --> ok
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12:01:24 How long does it take the larger wheel to reach 65 rpm?
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RESPONSE --> 65 rpm would equal an angular speed in rad / sec of (65/60) * 6.28 = 6.80 rad/sec 6.80 rad/sec / .58 rad/sec angular accel = 11.7 sec to reach 65 RPM
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12:01:40 ** 65 rpm is 65 * 2 `pi rad / min = 65 * 2 `pi rad / (60 sec) = 6.8 rad / sec, approx. At about .6 rad/s/s we get `dt = (change in ang vel) / (ang accel) = 6.8 rad / s / ( .6 rad / s^2) = 11 sec or so. **
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RESPONSE --> ok
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course phy 201 ???~s??^????`???z????assignment #030
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11:01:59 introductory set 8. If we know the constant moment of inertia of a rotating object and the constant net torque on the object, then how do we determine the angle through which it will rotate, starting from rest, in a given time interval?
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RESPONSE --> Know constant moment of inertia ""I"" Know net torque How much will it rotate from rest in given time period? net tau = I * alpha can be rearranged to alpha = net tau / I so dividing net torque by moment of inertia will give us the angular acceleration. having this we can then use the formula: vfinal = v0 + alpha * dt
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11:02:50 ** tau stands for torque and I stands for the moment of inertia. These quantities are analogous to force and mass. Just as F = m a, we have tau = I * alpha; i.e., torque = moment of inertia * angular acceleration. If we know the moment of inertia and the torque we can find the angular acceleration. If we multiply angular acceleration by time interval we get change in angular velocity. We add the change in angular velocity to the initial angular velocity to get the final angular velocity. In this case initial angular velocity is zero so final angular velocity is equal to the change in angular velocity. If we average initial velocity with final velocity then, if angular accel is constant, we get average angular velocity. In this case angular accel is constant and init vel is zero, so ave angular vel is half of final angular vel. When we multiply the average angular velocity by the time interval we get the angular displacement, i.e., the angle through which the object moves. **
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RESPONSE --> ok
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11:05:17 If we know the initial angular velocity of a rotating object, and if we know its angular velocity after a given time, then if we also know the net constant torque accelerating the object, how would we find its constant moment of inertia?
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RESPONSE --> moment of inertia = net torque / angular accel by taking the change in angular velocity / time we get angular accel. we are given the net torque, so we can plug these into the formula above to solve for moment of inertia.
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11:05:42 ** From init and final angular vel you find change in angular vel (`d`omega = `omegaf - `omega0). You can from this and the given time interval find Angular accel = change in angular vel / change in clock time. Then from the known torque and angular acceleration we find moment of intertia. tau = I * alpha so I = tau / alpha. **
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RESPONSE --> ok
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11:13:39 How do we find the moment of inertia of a concentric configuration of 3 uniform hoops, given the mass and radius of each?
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RESPONSE --> take the mass of each ring times the square of the radius from the point of rotation and total these three quantities.
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11:14:43 How do we find the moment of inertia a light beam to which are attached 3 masses, each of known mass and lying at a known distance from the axis of rotation?
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RESPONSE --> multiply each mass times the distance from the point of rotation and then add these three quantities. Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems.
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11:14:54 ** Moment of inertia of a mass r at distance r is m r^2. We would get a total of m1 r1^2 + m2 r2^2 + m3 r3^2. Note the similarity to the expression for the hoops. **
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RESPONSE --> ok
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11:17:19 Principles of Physics and General College Physics problem 8.4. Angular acceleration of blender blades slowing to rest from 6500 rmp in 3.0 seconds.
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RESPONSE --> blades drop from 6500 rpm to zero in 3sec. 6500 / 3 = - 2167 rpm/s
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11:18:12 The change in angular velocity from 6500 rpm to rest is -6500 rpm. This change occurs in 3.0 sec, so the average rate of change of angular velocity with respect to clock time is ave rate = change in angular velocity / change in clock time = -6500 rpm / (3.0 sec) = -2200 rpm / sec. This reasoning should be very clear from the definition of average rate of change. Symbolically the angular velocity changes from omega_0 = 6500 rpm to omega_f = 0, so the change in velocity is `dOmega = omega_f - omega_0 = 0 - 6500 rpm = -6500 rpm. This change occurs in time interval `dt = 3.0 sec. The average rate of change of angular velocity with respect to clock time is therefore ave rate = change in angular vel / change in clock time = `dOmega / `dt = (omega_f - omega_0) / `dt = (0 - 6500 rpm) / (3 sec) = -2200 rpm / sec. The unit rpm / sec is a perfectly valid unit for rate of change of angular velocity, however it is not the standard unit. The standard unit for angular velocity is the radian / second, and to put the answer into standard units we must express the change in angular velocity in radians / second. Since 1 revolution corresponds to an angular displacement of 2 pi radians, and since 60 seconds = 1 minute, it follows that 1 rpm = 1 revolution / minute = 2 pi radians / 60 second = pi/30 rad / sec. Thus our conversion factor between rpm and rad/sec is (pi/30 rad / sec) / (rpm) and our 2200 rpm / sec becomes angular acceleration = 2200 rpm / sec * (pi/30 rad / sec) / rpm = (2200 pi / 30) rad / sec^2 = 73 pi rad / sec^2, or about 210 rad / sec^2.
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RESPONSE --> ok
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11:23:58 Principles of Physics and General College Physics problem 8.16. Automobile engine slows from 4500 rpm to 1200 rpm in 2.5 sec. Assuming constant angular acceleration, what is the angular acceleration and how how many revolutions does the engine make in this time?
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RESPONSE --> initial speed = 4500 rpm final speed = 1200 rpm d speed = 3300 rpm dt = 2.5 sec angular accel? total number of revolutions? 3300 / 2.5 = - 1320 rpm/sec angular accel (4500 + 1200)/ 2 = 2850 rpm (171,000 rps) avg speed * 2.5 sec = 4.27 *10^5 revolutions
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11:31:06 The change in angular velocity is -3300 rpm, which occurs in 2.5 sec. So the angular acceleration is angular accel = rate of change of angular vel with respect to clock time = -3300 rpm / (2.5 sec) = 1300 rpm / sec. Converting to radians / sec this is about angular accel = -1300 rpm / sec ( pi / 30 rad/sec) / rpm = 43 pi rad/sec^2, approx.. Since angular acceleration is assumed constant, a graph of angular velocity vs. clock time will be linear so that the average angular velocity with be the average of the initial and final angular velocities: ave angular velocity = (4500 rpm + 1200 rpm) / 2 = 2750 rpm, so that the angular displacement is angular displacement = ave angular velocity * time interval = 2750 rpm * 2.5 sec = 6900 revolutions, approximately. In symbols, using the equations of uniformly accelerated motion, we could use the first equation `dTheta = (omega_0 + omega_f) / 2 * `dt = (4500 rpm + 1200 rpm) / 2 * (2.5 sec) = 6900 revolutions and the second equation omega_f = omega_0 + alpha * `dt, which is solved for alpha to get alpha = (omega_f - omega_0) / `dt = (4500 rpm - 1200 rpm) / (2.5 sec) = 1300 rpm / sec, which as before can be converted to about 43 pi rad/sec^2, or about 130 rad/sec^2. The angular displacement of 6900 revolutions can also be expressed in radians as 6900 rev = 6900 rev (2 pi rad / rev) = 13800 pi rad, or about 42,000 radians.
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RESPONSE --> I think there are a couple of errors here: first, (4500 + 1200) / 2 is 2850 rpm not 2750 next: I don't think we can multiply rpm's times dt in sec to get # of revolutions. Shouldn't it be 2850 rpm/60 = 47.5 rps * 2.5 sec = 118.75 total revolutions? I made an error in multiplying the rpm's times 60 instead of dividing when I converted them to rps.
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11:35:04 gen Problem 8.23: A 55 N force is applied to the side furthest from the hinges, on a door 74 cm wide. The force is applied at an angle of 45 degrees from the face of the door. Give your solution:
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RESPONSE --> the net tau = force * moment arm * sin theta (45deg) = 55N * .74m * .707 = 28.7 m*N
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11:36:05 ** ** If the force is exerted perpendicular to the face of the door, then the torque on the door is 55 N * .74 m = 40.7 m N. The rest of the given solution here is for a force applied at an angle of 60 degrees. You can easily adapt it to the question in the current edition, where the angle is 45 degrees:The torque on the door is 45 N * .84 m = 37.8 m N. If the force is at 60 deg to the face of the door then since the moment arm is along the fact of the door, the force component perpendicular to the moment arm is Fperp = 37.8 m N * sin(60 deg) = 32.7 N and the torque is torque = Fperp * moment arm = 32.7 N * .84 m = 27.5 m N. STUDENT COMMENT: Looks like I should have used the sin of the angle instead of the cosine. I was a little confused at which one to use. I had trouble visualizing the x and y coordinates in this situation.INSTRUCTOR RESPONSE: You can let either axis correspond to the plane of the door, but since the given angle is with the door and angles are measured from the x axis the natural choice would be to let the x axis be in the plane of the door. The force is therefore at 60 degrees to the x axis. We want the force component perpendicular to the door. The y direction is perpendicular to the door. So we use the sine of the 60 degree angle.
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RESPONSE --> ok
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11:45:40 gen problem 8.11 rpm of centrifuge if a particle 7 cm from the axis of rotation experiences 100,000 g's
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RESPONSE --> I was very confused by this one. Is the 100,000 g's the tangential accel? If yes, then setting the 100,000 g's = to r * (d angular velocity/dt) and rewriting it to solve for a tang should give me the answer but what is the dt?
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11:47:00 ** alpha = v^2 / r so v = `sqrt( alpha * r ) = `sqrt( 100,000 * 9.8 m/s^2 * .07 m) = `sqrt( 69,000 m^2 / s^2 ) = 260 m/s approx. Circumference of the circle is 2 `pi r = 2 `pi * .07 m = .43 m. 260 m/s / ( .43 m / rev) = 600 rev / sec. 600 rev / sec * ( 60 sec / min) = 36000 rev / min or 36000 rpm. All calculations are approximate. **
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RESPONSE --> So it was the centripetal accel that we wanted? not the tangential accel?
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11:56:55 gen problem 8.20 small wheel rad 2 cm in contact with 25 cm wheel, no slipping, small wheel accel at 7.2 rad/s^2. What is the angular acceleration of the larger wheel?
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RESPONSE --> The bigger wheel has a radius and, therefore, a cicumference which is 12.5 times as great as the small wheel. Therefore the large wheel's angular acceleration will be (7.2 rad/sec^2) / 12.5 = .576 rad /sec^2
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11:57:14 ** Since both wheels travel the same distances at the rim, angular displacements (which are equal to distance along the rim divided by radii) will be in inverse proportion to the radii. It follows that angular velocities and angular accelerations will also be in inverse proportion to radii. The angular acceleration of the second wheel will therefore be 2/25 that of the first, or 2/25 * 7.2 rad/s^2 = .58 rad/s^2 approx.. **
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RESPONSE --> ok
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12:01:24 How long does it take the larger wheel to reach 65 rpm?
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RESPONSE --> 65 rpm would equal an angular speed in rad / sec of (65/60) * 6.28 = 6.80 rad/sec 6.80 rad/sec / .58 rad/sec angular accel = 11.7 sec to reach 65 RPM
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12:01:40 ** 65 rpm is 65 * 2 `pi rad / min = 65 * 2 `pi rad / (60 sec) = 6.8 rad / sec, approx. At about .6 rad/s/s we get `dt = (change in ang vel) / (ang accel) = 6.8 rad / s / ( .6 rad / s^2) = 11 sec or so. **
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RESPONSE --> ok
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