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At t = 0 a grinding wheel has an angular velocity of 24.0 rad/s. It has a constant angular acceleration of 30.0rad/s2 until a circuit breaker trips at t = 2.00 s. From then on, it turns through 432 rad as it coasts to a stop at constant angular acceleration. (a) Through what total angle did the wheel turn between t = 0 and the time it stopped? (b) At what time did it stop? (c) What was its acceleration as it slowed down?

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Kazeemsodikisola

Answer:

θ=108rad

t =10.29seconds

α=-8.17rad/s²

Explanation:

Given that

At t=0, Wo=24rad/sec

Constant angular acceleration =30rad/s²

At t=2, θ=432rad as it try to stop because the circuit break

Angular motion

W=Wo+αt

θ=Wot+1/2αt²

W²=Wo²+2αθ

We need to find θ between 0sec to 2sec when the wheel stop

a. θ=Wot+1/2αt²

θ=24×2+1/2×30×2²

θ=48+60

θ=108rad.

b. W=Wo+αt

W=24+30×2

W=84rad/s

This is the final angular velocity which is the initial angular velocity when the wheel starts to decelerate.

Wo=84rad/sec

W=0rad/s, because the wheel stop at θ=432rad

Using W²=Wo²+2αθ

0²=84²+2×α×432

-84²=864α

α=-8.17rad/s²

It is negative because it is decelerating

Now, time taken for the wheel to stop

W=Wo+αt

0=84-8.17t

-84=-8.17t

Then t =10.29seconds.

a. θ=108rad

b. t =10.29seconds

c. α=-8.17rad/s²

A) The total angle θ the wheel turned between time of 0sec to when the wheel stopped is; θ = 108rad.

B) The time at which the wheel came to stop is; t = 10.29 seconds.

C) The acceleration at which it slowed down is; α = -8.17 rad/s²

We are given;

Initial time; t = 0 s

Initial angular velocity; ω₀ = 24rad/sec

Constant angular acceleration; α = 30rad/s²

time in motion; t = 2 s

Angle after time of 2 secs till it stops; θ = 432rad  

A) To find the total angle θ the wheel turned between time of 0sec to when the wheel stopped. Thus, we will use the formula;

 θ = ω₀t + ¹/₂αt²

plugging in the relevant values gives;

θ = (24 × 2) + (¹/₂ × 30 × 2²)

θ = 48 + 60

θ = 108rad.

B) To find the time that the wheel stopped, let us first find the final velocity at which the grinding wheel begins to decelerate using the formula;

ω = ω₀ + αt

plugging in the  relevant values gives;

ω = 24 + (30 × 2)

ω = 84 rad/s

Now, in deceleration to when it stops, it means;

ω₀ = 84rad/sec

ω  = 0rad/s

We can find the deceleration till it comes to a stop using the formula;

ω² = ω₀² + 2αθ

0² = 84² + (2 × α × 432)

-(84²) = 864α

α = -8.17 rad/s²

To find the time to come to rest;

ω = ω₀ + αt

0 = 84 - 8.17t

-84 = -8.17t

t = -84/-8.17

t = 10.29 seconds.

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