course PHY201 ???r?????????assignment #031????????????X?|?Physics I
......!!!!!!!!...................................
21:11:23 experiment to be viewed. What is the relationship between the angular velocity of the axle around which the string is wound and that of the large disk?
......!!!!!!!!...................................
RESPONSE --> The axle is the axis upon which the disk spins, and the disk is the rotating mass which will determine the amount of inertia dependent upon the radius of the disk. For angular velocity, all points on the object will rotate with the same angular velocity since every point moves throught the same angle in the same amount of time.
.................................................
......!!!!!!!!...................................
21:11:30 GOOD STUDENT RESPONSE The angular velocity of the axle and the angular velocity of the disk on the axle would be the same. However, the velocity would be different because they are of different distances from the center. In general, the axle will be moving at a slower speed(velocity) than a point on the outside of the disk. I am not sure if this is what you are asking. ** The speed of the falling object is the same as the speed of a point on the rim of the axle. The angular velocity of the axle is equal to the speed of a point on its rim divided by its radius: omega = v / r. The disk rotates with the axle so it has the same angular velocity. **
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
21:12:10 If the falling weight accelerates uniformly, does it follow that the rotating disk has a uniform angular acceleration?
......!!!!!!!!...................................
RESPONSE --> If the falling weight accelerates uniformly, it only makes sense that the disk will have a uniform angular acceleration as long as the radius of the disk remains constant.
.................................................
......!!!!!!!!...................................
21:12:14 GOOD STUDENT RESPONSE yes, because the angle of acceleration is proportional to the velocity of the disk with the radius(which is constant) as the constant of proportionality. And the velocity of the disk will be the same as the velocity of the falling weight which is dependent on the acceleration of the weight. ** If v changes at a uniform rate then since r is uniform, omega = v / r changes at a uniform rate. **
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
21:12:40 Principles of Physics and General College Physics Problem 8.28: Moment of inertia of bicycle wheel 66.7 cm diameter, mass 1.25 kg at rim and tire.
......!!!!!!!!...................................
RESPONSE --> To calculate the moment of inertia (I) we need to use the formula m * r^2 to find I. If the diameter is 66.7cm, then the radius will be .3315m. I = mr^2 I = (1.25kg)(.3315)^2 I = 0.137 kg * m^2
.................................................
......!!!!!!!!...................................
21:13:11 The mass of the rim and tire is all located at about the same distance from the axis of rotation, so the rim and tire contribute m * r^2 to the total moment of inertia, where m is the mass and r the distance from the axis of rotation of the rim and tire. The distance r is half the diameter, or 1/2 * 66.7 cm = 33.4 cm = .334 m, and the mass is given as 1.25 kg, so the moment of inertia of rim and tire is I = m r^2 = 1.25 kg * (.334 m)^2 = 1.4 kg m^2.
......!!!!!!!!...................................
RESPONSE --> I recalculated my answer and still got 0.137 kg * m^2.
.................................................
......!!!!!!!!...................................
21:13:29 Why can the mass of the hub be ignored?
......!!!!!!!!...................................
RESPONSE --> It is possible depending upon the radius of the hub.
.................................................
......!!!!!!!!...................................
21:13:45 The radius of the hub is less than 1/5 the radius of the tire; because its moment of inertia is m r^2, where r is its 'average' distance from the axis of rotation, its r^2 will be less than 1/25 as great as for the rim and tire. Even if the mass of the hub is comparable to that of the rim and tire, the 1/25 factor will make its contribution to the moment of inertia pretty much negligible.
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
21:17:27 gen Problem 8.38 arm, 3.6 kg ball accel at 7 m/s^2, triceps attachment 2.5 cm below pivot, ball 30 cm above pivot. Give your solution to the problem.
......!!!!!!!!...................................
RESPONSE --> To calculate the torque needed we will need to find Inertia and angular acceleration. Since the ball is 30cm above the point of axis: I = mr^2 I = (3.6kg)(.3m)^2 I = 0.324 kg * m^2 To find alpha, angular acceleration, we will use the formula: alpha = a/r alpha = 7m/s^2 / .3m alpha = 23.33 rad/s^2 now that we have these values we can calculate torque(tau). Tau = I * alpha Tau = (0.324 kg * m^2)(23.33 rad/s^2) Tau = 7.55 mN Next, we were asked to calculate the value of Force exerted by the tricep muscle. Since the muscle is 2.5cm from the point of axis, this value will be used for the value of r. And the tricep muscle is perpendicular to the forearm which is throwing the ball, so we must use 90 deg for angle theta. Tau = r * F * sin(theta) F = Tau / r * sin(theta) F = 7.55 mN / (.025m)(sin[90 deg]) F = 302N
.................................................
......!!!!!!!!...................................
21:17:39 ** The moment of inertia of a 3.6 kg ball at a point 30 cm from the axis of rotation is I = m r^2 = 3.6 kg * (.30 m)^2 = .324 kg m^2. At a 30 cm distance from axis of rotation the 7 m/s^2 acceleration becomes an angular acceleration of alpha = a / r = 7 m/s^2 / (.3 m) = 23.3 rad/s^2. The necessary torque is therefore tau = I * alpha = .324 kg m^2 * 23.3 rad/s^2 = 7.6 m N, approx.. The muscle exerts its force at a point x = 2.5 cm from the axis of rotation and perpendicular to that axis so we have F = tau / x = 7.6 m N / (.025 m) = 304 N. **
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
21:48:00 Univ. 10.52 (10.44 10th edition). 55 kg wheel .52 m diam ax pressed into wheel 160 N normal force mu =.60. 6.5 m N friction torque; crank handle .5 m long; bring to 120 rev/min in 9 sec; torque required? Force to maintain 120 rev/min? How long to coast to rest if ax removed?
......!!!!!!!!...................................
RESPONSE --> To calculate the torque required to accelerate a 55kg wheel to a angular velocity of 120 rev/min in 9s. First we need to find the velocity in the proper units. w = 120rpm * (2 pi)/ 60min) = 12.57 rad/s alpha = dw / dt = (12.57 rad/s - 0rad/s) / 9s alpha = 1.4 rad/s^2 Next we need to find inertia, and find the radius from the given diameter. Dia: .52m = r: .26m I = mr^2 I = (55kg)(.26m)^2 I = 3.72 kg * m^2 Tau = I * alpha Tau = (3.72 kg * m^2)(1.4 rad/s^2)^2 Tau = 7.29 mN To find the force required to maintain 120rpm, we must first calculate for net torque. Tau = 7.29mN Now we calculate torque friction Ffr = Fn * mu Ffr = 160N * .6 = 96N Tau = r * F Tau = (.26)(96N) = 24.96 mN Torque friction is -6.5 mN net Tau = T1 + T2 + T3 net Tau = 7.29 mN + 24.96 mN - 6.5 mN net Tau = 25.75 mN F = Tau / r F = 25.75 mN / .26m F = 99N If the ax is remove the friction torque will slow the wheel down. Tau = I * alpha alpha = Tau / I = (-6.5 mN)/(3.72 kg*m^2) alpha = -1.75 rad/s^2 alpha = dw / dt dt = dw / alpha = (0rad/s - 12.6 rad/s) / (-1.75 rad/s^2) dt = 7.2 sec
.................................................
......!!!!!!!!...................................
21:48:03 ** The system is brought from rest to a final angular velocity of 120 rev/min * 1min/60 sec * 2`pi/1 rev = 12.6 rad/s. The angular acceleration is therefore alpha = change in omega / change in t = 12.6 rad/s / (9 sec) = 1.4 rad/s^2, approx.. The wheel has moment of inertia I = .5 m r^2 = .5 * 55 kg * (.52 m)^2 = 7.5 kg m^2, approx.. To achieve the necessary angular acceleration we have tauNet = I * alpha = 7.5 kg m^2 * 1.4 rad/s^2 = 10.5 m N. The frictional force between ax and wheel is .60 * 160 N = 96 N at the rim of the wheel, resulting in torque tauFrictAx = -96 N * .52 m = -50 m N. The frictional torque of the wheel is in the direction opposite motion and is therefore tauFrict = -6.5 m N. The net torque is the sum of the torques exerted by the crank and friction: tauNet = tauFrictAx + tauFrict + tauCrank so that the torque necessary from the crank is tauCrank = tauNet - tauFrict - tauCrank = 10.5 m N - (-50 m N) - (-6.5 m N) = 67 m N. The crank is .5 m long; the force necessary to achive the 60.5 m N torque is therefore F = tau / x = 67 m N / (.5 m) = 134 N. If the ax is removed then the net torque is just the frictional torque -6.5 m N so angular acceleration is alpha = -6.5 m N / (7.5 kg m^2) = -.84 rad/s^2 approx. Starting at 120 rpm = 12.6 rad/s the time to come to rest will be `dt = `dOmega / alpha = -12.6 rad/s / (-.84 rad/s^2) = 14.5 sec, approx.. **
......!!!!!!!!...................................
RESPONSE -->
.................................................