1: Look at procedure 1. Suppose m =221 g (including the mass holder) is released

Question

1:

Look at procedure 1. Suppose m =221 g (including the mass holder) is released from rest. The pulley has diameter d = 4.74 cm of the smallest wheel (note: the string is wrapped around this smallest wheel in the 3-step pulley), and you measure the angular acceleration of the pulley = 0.488 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.
- g=9.80 m/s2

2)

Suppose the rod in procedure 2 has mass M =560 g, length L =77.8 cm, and rotates on a fixed, frictionless axis through the center of mass. Initially, assume there are no masses attached to the rod.

Assume two point masses of mass m = 718 g are attached to the rod, one on each side, at distance r = 17.6 cm from the center of mass of the rod. Find the moment of inertia of the rod-plus-masses system.
Irod-plus-masses = ______ kg-m2

1:

Look at procedure 1. Suppose m =221 g (including the mass holder) is released from rest. The pulley has diameter d = 4.74 cm of the smallest wheel (note: the string is wrapped around this smallest wheel in the 3-step pulley), and you measure the angular acceleration of the pulley = 0.488 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.
- g=9.80 m/s2

Calculate an absolute value of the torque exerted by the string on the pulley.
= ________ N-m

2)

Suppose the rod in procedure 2 has mass M =560 g, length L =77.8 cm, and rotates on a fixed, frictionless axis through the center of mass. Initially, assume there are no masses attached to the rod.

Assume two point masses of mass m = 718 g are attached to the rod, one on each side, at distance r = 17.6 cm from the center of mass of the rod. Find the moment of inertia of the rod-plus-masses system.
Irod-plus-masses = ______ kg-m2

Explanation / Answer

1)
Torque = rF
r is the radius of the pulley, r = 4.74/2
= 2.37 cm = 2.37 x 10-2 m

F = mg
m = 221 g = 0.221 kg
F = 0.221 x 9.8
= 2.1658 N

Torque = (2.37 x 10-2) x 2.1658
= 5.133 x 10-2 N. m

2)
Moment of inertia of the rod = ML2/12
M = 0.56 kg, L = 0.778 m
Irod = [0.56 x (0.778)2] / 12
= 0.02825 kg. m2.

Moment if inertia of the two masses
Im = 2 x mr2
= 2 x 0.718 x (0.176)2
= 0.04448 kg. m2.

Total moment of inertia = Irod + Im
= 0.02825 kg. m2 + 0.04448 kg. m2
= 0.7273 kg. m2.

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