A 2 kg mass is held at rest on top of a frictionless and horizontal table. A lig

Question

A 2 kg mass is held at rest on top of a frictionless and horizontal table. A light string loops over a pulley which is in the shape of a 10 cm radius solid disk which has a mass of 2 kg. The light string then supports a mass of 2 kg which is hanging in air

The mass on the table is released and the suspended mass falls. What is the acceleration of the falling mass.

What is the tension in the string which is attached to the sliding mass on the table?

What is the tension of the string which supports the hanging mass?

Explanation / Answer

The moment of intertia I for a disk is mr^2/2 = 0.01 = 1/100
Let alpha = angular acceleration of the disk
Torque = I*alpha = Fd*Rd where Fd is the force of the string on the disk and Rd = 0.1m
Fd is in the direction opposite to the weight of the hanging mass = g
Torque = 0.01*alpha = alpha/100 where alpha is the angular acceleration
The acceleration of the masses a is the same since they are connected by the string
The acceleration of the outside of the pulley = a = Rd*alpha => alpha = 10a
Torque = alpha/100 = 10a/100 = a/10 = Fd/10 => Fd = a
The net force Fn = g - Fd = g - a
The total mass being accelerated = 2kg
Fn = m*a = 2*a => a = Fn/2 = g/2 - a/2 => 3a/2 = g/2 => a = g/3 = 3.26m/s^2 <--------- A)

B) F = m*a = 2*a = 6.53N <----------------- B)

C) If the hanging mass were stationary, the tension in the string would be m*a = 9.8N
BUT the mass is accelerating downward 3.26m/s^2 so the tension will be
[9.8-3.26]*2kg = 11.76N <---------- C)
Notice that Fd = a/2 = 1.81N
11.76 - 1.81 = 9.95 N

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