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

Answer:

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

(a) is correct option.

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

$T\times r=I\times \alpha$

$T=\dfrac{I\times\alpha}{r}$

$T=\dfrac{\dfrac{mr^2}{2}\times\alpha}{r}$

$T=\dfrac{\dfrac{mr^2}{2}\times\dfrac{v}{r}}{r}$

Put the value into the formula

$T=\dfrac{\dfrac{125\times(0.2)^2}{2}\times a}{(0.1)^2}$

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

From equation (I) and (II)

$255a=5g$

Put the value into the formula

$a=\dfrac{5\times9.8}{255}$

$a=0.2\ m/s^2$

We need to calculate the  angular acceleration of the rod

Using formula of angular acceleration

$\alpha=\dfrac{a}{R}$

Put the value into the formula

$\alpha=\dfrac{0.2}{0.1}$

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

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