http://www.physics.qc.edu/files?453 (the procedures) Look at procedure 1. Suppos

Question

http://www.physics.qc.edu/files?453 (the procedures)


Look at procedure 1. Suppose m = 725 g (including the mass holder) is released from rest. The pulley has diameter d = 3.04 cm, and you measure the angular acceleration of the pulley ? = 0.985 rad/s2. Assume
- The system is frictionless.
- The string does not slip on the pulley.
- The string is always perpendicular to the diameter of the pulley.

a) Calculate a, the linear acceleration of m as it falls.
----------------m/s2

b) Calculate T, the tension in the string.
-------------N

c) Calculate ?, the torque exerted by the string on the pulley.
-----------N-m

d) Calculate I, the moment of inertia of the system.
-----------kg-m2


Suppose the rod in procedure 2 has mass M = 487 g, length L = 68.2 cm, and rotates on a fixed, frictionless axis through the center of mass. Initially, assume there are no masses attached to the rod.

a) Find the moment of inertia of the rod.
----------kg-m2

b) Now, assume two point masses of mass m = 763 g are attached to the rod, one on each side, at distance r = 25.8 cm from the center of mass of the rod. Find the moment of inertia of the rod-plus-masses system.
-----------------kg-m2

In each question: you place mass m on the mass holder, and released the system from rest. The system includes two pulleys and a long rod with two point masses on it.

a) In procedure 1: if you increase m, the mass on the mass holder, what happens to the following quantities:
- ?, the angular accelertion of the pulley.
- I, the moment of inertia of the pulleys-plus-rod system.
? decreases, I increases
? decreases, I decreases
? remains the same, I remains the same
? decreases, I remains the same
? remains the same, I decreases
? remains the same, I increases
? increases, I remains the same
? increases, I decreases
? increases, I increases

b) In procedure 2: if you move the two masses attached to the rotating rod away from the center of mass of the rod, what happens to the following quantities:
- ?, the angular accelertion of the pulley.
- I, the moment of inertia of the pulleys-plus-rod system.
? remains the same, I increases
? increases, I remains the same
? remains the same, I decreases
? decreases, I remains the same
? remains the same, I remains the same
? decreases, I decreases
? increases, I decreases
? increases, I increases
? decreases, I increases

Explanation / Answer

Procedure 1: We have, mg-T = ma and Torque = T*r = I*

But angular acceleration, = a/r

So, mg - Ia/r2 = ma

So, a = mg/(m+I/r2)

a) Putting values, acceleration a = r = (3.04/2)*0.985 = 1.5 cm/s2 = 0.015 m/s2

b) Tension T = mg-ma = m(g-a) = 0.725*(9.81-0.015) = 7.1 N

c) Torque = T*r = 7.1*(3.04/2) = 10.8 N-cm = 0.108 N-m

d) Moment of inertia I = / = 0.108/0.985 = 0.11 kg-m2

Procedure 2: Momnt of inertia of rod I = mL2/12 = 0.487*0.6822/12 = 0.019 kg-m2

Moment of inertia of rod + masses = I + 2*mr2 = 0.019 + 2*0.763*0.2582 = 0.12 kg-m2

a) When we increase mass m on placeholder,

- angular acceleration of the pulley increases

b) When we increase the masses,

- moment of inertia of the puley+rod system increases.

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