To determine the radius of gyration of a tire, an engineer lets the tire roll do

Question

To determine the radius of gyration of a tire, an engineer lets the tire roll down an inclined surface. If it takes the tire 3.0 seconds to roll 7.2 m (starting from rest) down an incline of 15degrees, what is the radius of gyration of the tire.

Radius of the tire is .33 m

Explanation / Answer

s=ut+1/2 *at^2 => 7.2 =0+ 0.5a*3^2 =>a= 1.6 m/s2 angular acceleration = a/r = 1.6/0.33 =4.848 rad/s2 now, mg sin15 -u*mg cos15 =ma => 2.536 -9.466 u =1.6 => coefficient of friction,u =0.099 torque is provided by frictional force. torque =I*angular acceleration => umg *cos15*r = m*k^2 *4.848 => radius of gyration, k = 0.253 m

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