Attached to each end of a thin steel rod of length 1.20 m and mass 6.60 kg is a

Question

Attached to each end of a thin steel rod of length 1.20 m and mass 6.60 kg is a small ball of mass 1.09 kg. The rod is constrained to rotate in a horizontal plane about a vertical axis through its midpoint. At a certain instant, it is rotating at 39.0 rev/s. Because of friction, it slows to a stop in 32.0 s. Assume a constant frictional torque.

(b) Compute the retarding torque due to friction.

(c) Compute the total energy transferred from mechanical energy to thermal energy by friction.

(d) Compute the number of revolutions rotated during the 32.0 s.

(e) Now suppose that the frictional torque is known not to be constant. Which, if any, of the quantities can still be computed without additional information?

none of these

number of revolutions    

angular acceleration

mechanical energy loss

retarding torque

If such a quantity exists, give its value.

Explanation / Answer

(b) wf = 0

wi = 39 rev/s = 39 x 2pi rad/s = 245 rad/s

wf = wi + alpha t

alpha =7.66 rad/s^2

torque = I alpha

I = (6.60 x 1.20^2 / 12) + ( 2 x 1.09 x (1.20/2)^2)

I = 1.5768 kg m^2

torque = 1.5768 x 7.66 = 12.08 N m

(c) Energy transformed = I w^2 /2 - 0

= 1.5768 x 245^2 / 2

= 47323.71 J

(d) wf^2 - wi^2 = 2 alpha theta

0^2 - 245^2 = 2(-7.66) (theta)

theta = 3918 rad

revolutions = 3918 / 2pi =623.5 rev

(e) Mechnical energy loss

{ because it only depends on final and initial angular speed}

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