q9.5 At a time t = 3.20 s , a point on the rim of a wheel with a radius of 0.160

Question

q9.5

At a time t = 3.20 s , a point on the rim of a wheel with a radius of 0.160 m has a tangential speed of 51.0 m/sas the wheel slows down with a tangential acceleration of constant magnitude 10.7 m/s2 .

Part A

Calculate the wheel's constant angular acceleration.

Part B

Calculate the angular velocity at t = 3.20 s .

Part C

Calculate the angular velocity at t=0.

Part D

Through what angle did the wheel turn between t=0 and t = 3.20 s ?

Part E

Prior to the wheel coming to rest, at what time will the radial acceleration at a point on the rim equal g = 9.81 m/s2 ?

Explanation / Answer

part(A)

angular acceleration = alfa = -atan/R = -10.7/0.16 = -66.875 rad/s^2

part(B)

angular velocity = w = v/R

w = 51/0.16 = 318.75 rad/s

part(C)

w = wo + alfa*t

wo = angular speed at t = 0

318.75 = wo - (66.875*3.2)

wo = 318.75+(66.875*3.2)

wo = 532.75 rad/s

--------

part(D)

angular displacement = theta = wo*t + 0.5*alfa*t^2

theta = (532.75*3.2)-(0.5*66.875*3.2^2) = 1362.4 radians

---------------

part(E)

arad = R*w^2

w = sqrt(a/R)

w = sqrt(9.81/0.16) = 7.83 m/s

from equation of motion

w = wo + alfa*t

7.83 = 532.75-(66.875*t)

t = 7.85 s   <------answer

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