A wheel of mass 10 kg and radius 2m rotates with angular velocity w= 2pi rad/s

Question

a)If constant torque is applied to the wheel and it comes to rest in 2 seconds, what is the magnitude of the angular acceleration?

b) Calculate the magnitude of the torque required to bring the wheel to rest in 2 seconds? Assume wheel is solid disk.

c)What is the change in kinetic energy of the wheel?

d) What is the work done by the external torque?

e) If instead, we assume that the wheel is a hoop with the same mass and radius, will the magnitude of the torque required to stop be less than, greater than, or the same as the torque calculated in part b? Explain.

Answer 1

(a)

 

α = |Δω/Δt| = |[(0 - 2π) rad/s]/(2s)| = π rad/s²

 

(b)

 

τ = Iα = (mR²/2)α = [(10)(2)²/2]π Nm = 20π Nm

 

(c)

 

ΔK = I/2 (ω_f² - ω_i²) = (1/2) [(10)(2)²/2] [0² - (2π)²] J = -40π² J

 

(d)

 

W = ΔK = -40π² J

 

(e)

 

I changes from mR²/2 to mR², which is twice the I in parts (a)-(d).  Since

 

τ = Iα,

 

the required torque will be double for the same angular acceleration.