Question

A mass of 5.00 kg pulls down vertically on a string that is wound around a rod of radius 0.100 m and negligible moment of inertia. The rod is fixed in the center of a disk. The disk has mass 125 kg and radius 0.2 m. They turn freely about a fixed axis through the center. What is the angular acceleration of the rod, in radians/s?

a. 2
b. 25
c. 50
d. 75
e. 125

Answer 1

The angular acceleration of the rod is 2 rad/s².

(a) is correct option.

Step-by-step explanation:

Given that,

Mass of string = 5.00 kg

Radius = 0.100 m

Mass of disk = 125 kg

Radius of disk = 0.2 m

We need to calculate the acceleration

Using balance equation

`mg-T=ma`

Put the value of m

`5g-T=5a`....(I)

We need to calculate the tension

Using balance equation

`T* r=I* \alpha`

`T=(I*\alpha)/(r)`

`T=((mr^2)/(2)*\alpha)/(r)`

`T=((mr^2)/(2)*(v)/(r))/(r)`

Put the value into the formula

`T=((125*(0.2)^2)/(2)* a)/((0.1)^2)`

`T=250a`....(II)

From equation (I) and (II)

`255a=5g`

Put the value into the formula

`a=(5*9.8)/(255)`

`a=0.2\ m/s^2`

We need to calculate the angular acceleration of the rod

Using formula of angular acceleration

`\alpha=(a)/(R)`

Put the value into the formula

`\alpha=(0.2)/(0.1)`

`\alpha=2\ rad/s^2`

Hence, The angular acceleration of the rod is 2 rad/s².