A wheel of radius R = 2 m is initially rotating at w = 3 rad/sec. The wheel is s

Question

A wheel of radius R = 2 m is initially rotating at w = 3 rad/sec. The wheel is slowed down by a machine at a deceleration rate w = -.25 rad/s2 a) How many seconds does it take the wheel to come to full stop? b) How many turns does it make before coming to full stop? c) What is the tangential velocity at the edge of the wheel 1 second after deceleration? d) What is the tangential acceleration at the edge of the wheel? e) What is the centripetal acceleration at the edge of the wheel 1 second after deceleration?

Explanation / Answer

For part A. just use the formula: a=(vf-vi)/t where Vi is initial and Vf is final velocity. a=-0.25 rad/s^2 t=? Vi=3 rad/s Vf=o rad/s So -0.25 rad/s^2=(0-3)/t t=-3/-0.25 t=12 seconds Part B. Use the equation d=1/2at^2+Vit d=1/2*(-0.25)(12^2)+3*12 d=-0.125*144+36 d=-18+36 d=36 radians Part C. Use v=wr where w is the angular velocity, v is the tangential velocity, and r is radius Now it says 1s later so lets find the angular velocity after 1 second. So -0.25=(Vf-3)/1s Vf=2.75 rad/s So using the equation above, v=2m * 2.75 rad/s So v=5.5 m*rad/s or you could say 5.5 m/s Part D. Use equation a=v^2/r where a is tangential acceleration and v is tangential velocity. a=3^2/2 so a=9/2 a=4.5 m/s^2 Part E. Dont know how to solve centripetal acceleration. But I think it is a=w^2*r So a=(2.75)^2*3 a=22.69 rad/s^2 Please Rate Answer!! :)

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