Question

A 1.1 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 1.1 kg. The light string then supports a mass of 1.1 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?

Answer 1

T1 = tension of the string which supports the hanging mass.
T2 = tension in the string which is attached to the sliding mass.

The hanging object has 2 vertical forces. The weight is causing it to accelerate as the tension in the string is causing it to accelerate.

Net downward force = Weight – Tension = mass * a
=> 1.1 * 9.8 – T1 = 1.1 * a
=> T1 = 10.78 – 1.1 * a                                      ..............................(1)


The tension in the string attached to the sliding object is causing it to accelerate.
T2 = 1.1 * a                           .......................................................... (2)

The pulley has 2 forces, which produce the net torque, which are affecting its angular acceleration.
Torque = Force * radius in meters = F * 0.1

The torque produced by the tension in the string which supports the hanging mass is causing the pulley to accelerate. The torque caused by the tension in the string which is attached to the sliding mass is causing the pulley to decelerate.

Torque produced by T1 = T1 * 0.1
Torque produced by T2 = T2 * 0.1

Net torque = T1 * 0.1 – T2 * 0.1 = (T2 – T1) * 0.1

AND

Net torque = moment of inertia * angular acceleration
Mmoment of inertia = 1/2 * m * r² = 1/2 * 1.1 * 0.1² = 0.0055 kgm²
Angular acceleration = linear acceleration ÷ radius = a ÷ 0.1


Net torque = Moment of inertia * Angular acceleration = 0.0055 * (a ÷ 0.1) = a * 0.055
Net torque = a * 0.055

(T2 – T1) * 0.1 = a * 0.055
T2 – T1 = a * .55
put eq (1) and eq (2)
=> 10.78 – 1.1 * a – 1.1 * a = .55 * a
=> 10.78 – 2.2 * a = .55 * a
=> 10.78 = 2.75 * a
=> a = 3.92 m/s²

T1 = 10.78 – 2 * 3.92 = 2.94 N
T2 = 2 * 3.267 = 7.84 N