A toy Ferris Wheel undergoes an angular acceleration such that the angular speed

Question

A toy Ferris Wheel undergoes an angular acceleration such that the angular speed goes from zero to 2.89 rad/s in 1.33 s. Assume that the Ferris Wheel is a circular disk with radius 0.29 m and mass 6.7 kg. What is the magnitude of the net torque on the Wheel in N A toy Ferris Wheel undergoes an angular acceleration such that the angular speed goes from zero to 2.89 rad/s in 1.33 s. Assume that the Ferris Wheel is a circular disk with radius 0.29 m and mass 6.7 kg. What is the magnitude of the net torque on the Wheel in N

Explanation / Answer

Here ,

wi = 0 rad/s

wf = 2.89 rad/s

Now , using first equation of motion

wf = wi + a*t

a = (2.89 - 0)/1.33

a = 2.17 rad/s^2

Now , as moment of inertia , I = 0.5 * 6.7 * 0.29^2

I = 0.282 Kg.m^2

Using second law of motion

T = I*a

T = 0.282 * 2.17

T = 0.611 N.m

the magnitude of net torque acting on the wheel is 0.611 N.m

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