Question

A child abuses an antique record player to further his knowledge of physics. The record turntable can be treated as a uniform disk IMR2) with mass 0.75 kg and radius 0.16 m that spins freely.
First, the child wraps a long string several times around the turntable's circumference. He then pulls horizontally on the string with a constant linear acceleration of 1.2 m/s2, causing the turntable to begin rotating from rest. Assuming that the string doesn't slip, what angular speed will the turntable have once the child has pulled the end of the string a distance of 2.0 m?
Next, the child drops a handful of spaghetti onto the turntable (the spaghetti sticks to the turntable). The child then observes that the rotation rate of the turntable is now 5.0 rad/s slower. Find the moment of inertia of the spaghetti about its rotation axis. If you didn't answer the previous part, use a value of 12 rad/s for the initial angular velocity of the turntable (so it ends up at 7.0 rad/s)
To stop the spinning mess, the child could press his thumb against the turntable near the rotation axis or near the outer edge of the turntable. Which choice would result in the child needing to do the LEAST amount of work to stop the turntable, or is it a tie? Justify your answer with a conceptual argument or a calculation

Answer 1

Answer:

(a)7.5  rad/s2

(b)1.83s

(c)0.03 kgm2

(d)At the outer edge

Explanation:

The moments of inertia of the record table can be calculated as:

$I = MR^2 = 0.75*0.16^2 = 0.0192kgm^2$

(a) If he is pulling with a constant linear acceleration of a= 1.2 m/s2, then the constant angular acceleration is

$\alpha = \frac{a}{R} = \frac{1.2}{0.16} = 7.5 rad/s^2$

(b)The time it takes for the child to pull a distance of s = 2m given a = 1.2 m/s2

$s = \frac{at^2}{2}$

$t^2 = \frac{2s}{a} = \frac{2*2}{1.2} = 3.33$

$t = 1.83s$

Then the angular speed the turntable would have achieved by that time is

$\omega = \alpha t = 7.5*1.83 = 13.7rad/s$

(c) By the law of conservation in angular momentum:

$I_1\omega_1 = I_2\omega_2$

where I1 is the initial moment of the turn table before spaghetti drop, and I2 is after.

$0.0192*13.7 = I_2(13.7 - 5)$

$I_2 = \frac{0.0192*13.7}{8.7} = 0.03 kgm^2$

(d) For the same force, the child could generate different amount of torque, depending on where he's pressing his thumb. If it's near the the rotational axis, the moment arm is very small, or not at all, making the torque small. If it's at the edge, then the moment arm is large, making greater torque, so less work.