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

Answer:

The no of revolutions is 2.032 revolution.

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

$\tau=I\times\alpha$

$\alpha=\dfrac{\tau}{I}$

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

Where, F = force

r = radius

I = moment of inertia

Put the value into the formula

$\alpha=\dfrac{32\times170\times10^{-3}}{0.85}$

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

We need to calculate the rotational speed

Using equation of angular motion

$\omega_{f}=\omega_{i}+\alpha t$

$\omega_{f}=6.4\times2$

$\omega=12.8\ rad/s$

We need to calculate the angular position

Using equation of angular motion

$\theta=\omega_{i}+\dfrac{1}{2}\alpha t^2$

$\theta=0+\dfrac{1}{2}\times6.4\times4$

$\theta=12.8\ radian$

We need to calculate no of revolutions

$n = \dfrac{\theta}{2\pi}$

$n=\dfrac{12.8}{2\times3.15}$

$n=2.032\ revolution$

Hence, The no of revolutions is 2.032 revolution.