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course Phy201
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course Phy201
I am confused on a few questions but here are the ones that I attempted
experiments conducted today.#$&*
For the experiment where the domino slipped off the strap if it was going too fast, what was the maximum angular velocity of the system for which the domino did not slip off?
179 deg/s
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How far was the domino from the axis of rotation?
13cm
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How fast was the domino therefore moving?
Vf = 358deg/s
358deg/360deg*13cm = 12.9 cm/s
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358 deg / s would be very nearly 1 revolution per second.
The arc distance for a complete revolution is much greater than the radius of the circle. So the speed of the domino would be a lot more than 12.9 cm/s.
If you did 358/360 of the circumefernce of the circle, you would get the arc distance traveled in 1 second.
The Arc distance traveled was 0.99 (what untis)
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The centripetal acceleration of an object moving with speed v around a circle of radius r is
a_centripetal = v^2 / r.
The centripetal acceleration of the domino is the acceleration toward the center required to keep it moving in a circular path. What was the maximum centripetal acceleration, among your trials, for which the domino did not slip off?
acent = 12.9m/s^2(13cm)= 12.8cm/s^2
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With the correct velocity (see above) this calculation would give you the correct acceleration.
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What was the coefficient of friction between domino and strap, based on the slope required for the domino to begin sliding?
8cm/28.9cm = 0.28
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Assuming that the ball in the experiment required .42 seconds to fall to the floor, what speeds do you conclude for each trial?
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The free fall point was used as our standard point.
When the ramp distance was 70 cm with ½ mag from the edge the ball landed 24 cm from the free fall point.
70 cm ramp no mag the ball fell 26 cm from free fall point.
70cm ramp and all the mag on the ramp 24.5 cm from the free fall point.
20cm ramp and no mag the ball fell 13.5 cm from the free fall point.
20cm ramp with all the mag on the ramp the ball fell 14.5 cm from the free fall point.
26cm/.42s = 61.9cm/s
24cm/.42s = 57.0cm/s
24.5cm/.42s = 58.3cm/s
13.5cm/.42s = 32.1cm/s
14.5cm/.42s = 34.5cm/s
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You don't seem to be distinguishing between the displacement of the ball from the free fall point and the displacement perpendicular to the original line. I believe the shorter distances (e.g., 13.5 cm) correspond to distances from the original path, or perhaps from the landing point of the undeflected ball, but I don't think this is the distance from the free fall point.
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Do you conclude that the magnet sped the ball up, slowed it down, or that it had no significant effect?
The mag may have sped up the ball but not enough change to measure without timing it
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For each set of trials, how much speed did the magnet induce in the direction perpendicular to the original direction?
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The may induced a small amount of speed on the direction perpendicular but not enough to greatly effect the original.
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The magnet is 5 cm long. Assuming that the force exerted by the magnet has a significant effect for only this distance, for what time interval was the ball influenced by the magnet?
The time interval when it was close enough to be drawn into the magnetic field and the force stayed on the ball until it dropped off the edge.
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By how much did the momentum of the ball change due to the effect of the magnet, based on the velocity it attained perpendicular to its original line of motion? Assume the ball's mass to be 20 grams.
Mom = mv
I know the mass = 0.2kg but how do I find the velocity if I do not know the time that it took to displace the ball from its original free fall position???
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You need to base this on the distance perpendicular to the original line. That deflection is the result of the magnet's effect.
For example if the magnet deflected the ball 13.5 cm from the original path, the momentum change in that direction would be based on the 32.1 cm/s velocity.
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What do you conclude was the average force exerted on the ball by the magnet?
I need the time to do this also or am I forgetting to do something else to find it
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You were asked previously how much time the ball spent in front of the magnet. You know how long the magnet is and you previously calculated how fast it was traveling at the end of the ramp. So you should be able to find the time during which the magnetic force resulted in the momentum change.
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`q002. The magnet and domino on the strap were located 10.5 cm and 17 cm from the axis of rotation. On one trial the system rotated through 210 degrees in 6 seconds, ending up at rest.
What were the average and initial angular velocities of the system, in degrees / second, if we assume uniform angular acceleration?
angv = 210deg/6s = 35deg/s
initial v = 70deg/s
final v = 0deg/s
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What is the circumference of each of the circles around which the magnet and domino traveled?
C = 2pi(r)
C mag = 2pi(10.5cm) = 66cm
C dom = 2pi(17cm) = 107cm
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How far did each actually travel during the 210 degree rotation?
Magnet = 210deg/360deg*66cm = 39cm
Domino = 210deg/360deg*107cm = 62cm
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What therefore was the average speed of each around the arc, in units of distance / time?
210deg/6s = 35deg/s
Mag = (36deg/s)/360deg*66cm = 6.4cm/s
Dom = (36deg/s)/360deg*107cm = 10.4cm/s
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What was the initial speed of each?
Mag = 70/360*66cm = 13cm/s
Dom = 70/360*107 = 21cm/s
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`q003. A radian is the angle subtended by the arc of a circle whose arc length is equal to the radius of that circle.
What are the lengths of the arcs of a circle of radius 15 cm corresponding to angles of 2 radians, 1/2 radian and 6 radians?
2rad = 30cm
½ rad = 7.5cm
6rad = 90cm
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What are the angles subtended on the same circle by arcs of 45 cm, 5 cm and 25 cm?
2 rad = 30cm
½ rad = 7.5cm
6rad = 90cm
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Previously you found that 2 radians of rotation corresponded to 30 cm of arc, and 6 radians to 90 cm of arc.
Now you have 45 cm of arc. To how many radians of rotation would that correspond for this circle?
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What is the angle subtended by an arc consisting of the entire circumference of the circle?
45cm - 3rad
5cm = 1/3 rad
25cm = 5/3rad
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The angle subtended by an arc consisting of the entire circumference of any circle is 360 degrees. That angle is also 2 pi radians.
How many degrees are therefore contained in a radian?
57.3deg
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How many radians are there in a degree?
About 0.012rad
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Returning to the preceding problem, where the strap rotated through 210 degrees in 6 seconds, through how many radians did the strap rotate? 210deg/360deg*2pi15 = 3.66rad
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Based on your result for the number of radians, what were the average and initial angular velocities of the strap?
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Initial 70deg/s or 1.22 rad/s
Ave = 35deg/s or 0.61rad/s
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Your answers to the preceding question would be in radians / second. If the initial angular velocity was maintained, through how many radians would the system rotate in one second?
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1.22 rad/s
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Recall that one radian of angle corresponds to an arc distance equal to the radius. What arc distance would therefore correspond to one second's worth of rotation for the domino, were the initial angular velocity to be maintained?
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1.22rad/s*15cm = 18.3cm*rad/s
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What arc distance would correspond to one second's worth of rotation for the magnet, were the initial angular velocity to be maintained?
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70deg/s
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That is the angular velocity in degrees / second, but it's not a distance. Distance is measured in units of length (e.g., centimeters).
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How fast are the domino and the magnet therefore moving at the initial instant?
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70deg/360deg*15cm = 3cm/s
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15 cm is not the arc distance that corresponds to 70 degrees.
It should be clear just from a sketch that a 70 degree arc on a circle of radius 15 cm will be a lot longer than 3 cm.
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`q004. Assume that the center of the strap is halfway between the domino and the magnet. How far then is the center of the strap from the axis of rotation?
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3cm
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If the strap has mass 70 grams then what is its torque about the axis of rotation? The torque is equal to the weight of the strap multiplied by the distance of its center from the axis of rotation.
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T= weight*distance of its center from the axis of rotation
T = 70g*3cm = 210gcm
Or would this be in Ncm???
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Good question.
You multiplied the mass of the strap by the distance, not the weight.
What is the weight of the strap, and what therefore is the torque?
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If the domino has mass 17 grams then what is its torque about the axis of rotation?
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T = 17g*3cm = 51gcm
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The domino is not 3 cm from the axis.
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Are the domino and the center of strap both on the same side of the axis of rotation, or on opposite sides? Do their torques therefore reinforce or work counter to one another?
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Yes same side
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What therefore is the total torque exerted by strap and domino?
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210gcm+51gcm = 261 gcm
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The torque produced by the magnet is equal and opposite to the sum of the other torques. What therefore is the magnitude of that torque, and what do you conclude is the mass of the magnet?
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M1*d1-m2*d2 = 0
17g*17cm-m2*10.5cm =0
M2=27.5grams
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You're on the right track on this problem but you need to rework those torques.
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University Physics students: Find the moment of inertia of the strap itself, assuming it to be a uniform rod of length 30 cm and mass 70 grams. Note that the strap doesn't rotate about its center, so you'll have to do an integral. Begin by partitioning the length of the strap.
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`q005. If you got a reasonable result for the mass of the magnet, use it. Otherwise assume that the mass of the magnet is 50 grams.
You previously determined how fast the magnet and the domino were moving at the initial instant. What therefore were their kinetic energies?
KE = 1/2mv^2
KE = ½(17g)(10cm/s)^2
KE = 850g*(cm/s)^2
KE = ½(27.5g)(7cm/s)^2
KE = 673.75g*(cm/s)^2
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The magnet is about three times as massive as the domino, both are moving at the same angular velocity, but the kinetic energy of the magnet is much less than three times the kinetic energy of the domino. How can this be?
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Our measurements were smaller but not as much as 3 times smaller unless my calculations are wrong
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You'll want to finish working this, and in the process you'll want to rework some of your existing questions. Check my notes.
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I'm not sure 10 cm/s and 7 cm/s correspond to the initial velocities, but this is how you would calculate the kinetic energies.
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&#Good work. See my notes and let me know if you have questions. &#