Question

A pulley has a moment of inertial of 0.85kg m about an axle and a radius of 170mm. The string is wrapped around the pulley exerts a constant force of magnitude 32N. Determine the angular acceleration of the pulley. Find the rotational speed of the pulley at t = 2s. How many revolutions did the pulley make during this time?

Answer 1

The no of revolutions is 2.032 revolution.

Step-by-step explanation:

Given that,

Moment of inertia = 0.85 Kgm²

Radius = 170 mm

Force = 32 N

Time = 2s

We need to calculate the angular acceleration

Using formula of torque

`\tau=I*\alpha`

`\alpha=(\tau)/(I)`

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

Where, F = force

r = radius

I = moment of inertia

Put the value into the formula

`\alpha=(32*170*10^(-3))/(0.85)`

`\alpha=6.4\ m/s^2`

We need to calculate the rotational speed

Using equation of angular motion

`\omega_(f)=\omega_(i)+\alpha t`

`\omega_(f)=6.4*2`

`\omega=12.8\ rad/s`

We need to calculate the angular position

Using equation of angular motion

`\theta=\omega_(i)+(1)/(2)\alpha t^2`

`\theta=0+(1)/(2)*6.4*4`

`\theta=12.8\ radian`

We need to calculate no of revolutions

`n = (\theta)/(2\pi)`

`n=(12.8)/(2*3.15)`

`n=2.032\ revolution`

Hence, The no of revolutions is 2.032 revolution.