Physics, 26.02.2020 02:50, thebrain1345

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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answered: ianocampo3413

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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At t = 0 a grinding wheel has an angular velocity of 24.0 rad/s. It has a constant angular accelerat...