2. A merrygoround 5 meters in diameter accelerates uniformly from 1 rpm to 30 rp

Question

2. A merrygoround 5 meters in diameter accelerates uniformly from 1 rpm to 30 rpm in 10 seconds.
a. what is the angular acceleration of the wheel
b. How far will a point on the edge of the merrygoround travel in this time?
c. Assuming angular acceleration is constant, how much additional time does it take the merrygoround to reach 60 rpm?
d. The moment of inertia of the merrygoround is 1.5 x 10^7 kg m^2, calculate the torque required to accelerate the merrygoround.
e. What is the kinetic energy of the merrygoround while spinning at 60 rpm.

Please show work, need to replicate on a test

Explanation / Answer

here,

diameter = 5 m

radius , r = 2.5 m

initial angular velocity , Wi = 1 rpm

Wi = 1 * 2 * pi / 60

Wi = 0.104 rad/s

final angular velocity , Wf = 30 rpm

Wf = 30 * 2 * pi / 60

Wf = 3.14 rad/s

time taken ,t = 10 s

a)

using first equation of motion

Wf = Wi + a * t

3.14 = 0.104 + a * 10

a = 0.303 rad/s^2

angular accelration of the wheel is 0.303 rad/s^2

b)

let the distance travelled be theta

using seccond equation of motion

theta = Wi * t + 0.5 * a * t^2

theta = 0.104 * 10 + 0.5 * 0.303 * 100

theta = 16.19 rad

distance = theta * r

distance = 2.5 * 16.19

the edge of merrygoround travels 40.47 m

c)

let final speed , Wf = 60 rpm

Wf = 60 * 2 * pi /60

Wf = 6.28 rad/s

using first equation of motion

Wf = Wi + a * (t + 10)

6.28 = 0.104 + 0.303 * (t+10)

t = 10.3 s

additional time taken is 10.3 s

d)

torque = I * a

torque = 1.5 * 10^7 * 0.303

torue required is 4.54 * 10^6 N.m

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