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kicyunya [14]
3 years ago
7

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 exert

s 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?
Physics
1 answer:
Dmitriy789 [7]3 years ago
7 0

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.

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