**Unformatted text preview: **2 k rod = °-° + =-+ = θ μ θ F ω F b) With respect to the center of the roll, the rod and the normal force exert zero torque. The magnitude of the net torque is n μ f R f F k and , ) ( =-may be found insertion of the value found for rod F into either of the above relations; i.e., N. 2 . 33 sin rod k = = θ F μ f Then, . rad/s 71 . 4 ) m kg (0.260 m) 10 N)(18.0 54 . 31 N . 40 ( 2 2 2 = ⋅ ×-= =-I τ α...

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- Force, 2 m, 9.80 m/s, 0.260 kg, 4.71 rad, 16.0 kg