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Starting from rest, a disk rotates about its central axis with constant angular acceleration. In 5.00 s, it rotates 10.1 rad. During that time, what are the magnitudes of (a) the angular acceleration and (b) the average angular velocity? (c) What is the instantaneous angular velocity of the disk at the end of the 5.00 s? (d) With the angular acceleration unchanged, through what additional angle (rad) will the disk turn during the next 5.00 s?

User Nshoo
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1 Answer

4 votes

Answer:

a

The angular acceleration is
\alpha = 0.808 rad/s^2

b

The average angular velocity is
w = 2.02 \ rad/s

c

The instantaneous angular velocity is
w = 4.04 \ rad/s

d

The additional angle (rad) will the disk turn during the next 5.00 s is


\theta_n = 30.30 \ rad

Step-by-step explanation:

From the question we are told that

The duration of rotation is
t = 5.00s

The angular displacement is
\theta= 10.1 rad

From newtons Law the angular displacement is mathematically represented as


\theta = w_o t +(1)/(2) \alpha t^2

Where
\alpha is the angular acceleration


w_o is the initial angular velocity and its value is 0

Making
\alpha the subject of formula


\alpha = (\theta)/(0.5 * t^2)

Substituting value


\alpha = (10.1)/(0.5 * 5^2 )


\alpha = 0.808 rad/s^2

The average angular velocity is mathematically represented as


w = (\theta )/(t)

Substituting value


w = (10.1)/(5)


w = 2.02 \ rad/s

From the equations of motion the instantaneous angular velocity is mathematically represented as


w = w_o + \alpha t

Substituting value


w = 0 + 0.808 * 5


w = 4.04 \ rad/s

Modifying the equation for angular displacement is mathematically represented as


\theta_(n) = w_o t + 0.5 \alpha t^2

From the question
\alpha is not changed

t = 5s

the initial velocity would be the instantaneous velocity

So
w_o = 4.04 \ rad/s

Substituting value


\theta_n = 4.04 * 5 + (0.5 * 0.808 * 5^2)


\theta_n = 30.30 \ rad

User Sohail
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