Rolling Cylinder and the Static-Friction Limit
A uniform solid cylinder of mass 2.0 kg and radius 0.10 m is released from rest on a 30° incline. The coefficient of static friction is 0.15 and the coefficient of kinetic friction is 0.10. The cylinder’s axis is horizontal, and it is free to rotate about that axis. A student assumes it rolls without slipping and calculates the acceleration using only the component of gravity along the incline. Take g=9.8 m/s². Compare the static friction required for rolling without slipping with the available limit before deciding whether that assumption is valid.
1. Assuming rolling without slipping, find the cylinder’s acceleration and the required static friction. Then check whether the available static friction is sufficient.
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For a solid cylinder, I = 0.5 M R². Let friction f point up the incline. Newton’s second law along the incline: M g sinθ − f = M a. Torque about the center: f R = I α = (0.5 M R²)(a/R), so f = 0.5 M a.
Substitute: M g sinθ − 0.5 M a = M a, so a = g sinθ/1.5 = (9.8)(0.5)/1.5 = 3.27 m/s². Then f = 0.5(2.0)(3.27) = 3.27 N.
The normal force is N = M g cosθ = (2.0)(9.8)(0.866) = 16.97 N. The required coefficient is μ_req = f/N = 3.27/16.97 = 0.192. Since 0.192 > 0.15, the available static friction is insufficient; the cylinder cannot roll without slipping.
2. Given that the cylinder slips, find its linear acceleration and angular acceleration using kinetic friction.
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With slipping, kinetic friction acts up the incline: f_k = μ_k N = (0.10)(16.97) = 1.70 N.
Linear motion: M g sinθ − f_k = M a, so a = [(2.0)(9.8)(0.5) − 1.70]/2.0 = (9.8 − 1.70)/2.0 = 4.05 m/s².
Angular motion: f_k R = I α = 0.5 M R² α, so α = 2 f_k/(M R) = 2(1.70)/[(2.0)(0.10)] = 17.0 rad/s².
Check: Rα = (0.10)(17.0) = 1.70 m/s², which is less than a = 4.05 m/s². Thus the contact point slips down the incline relative to the surface, consistent with kinetic friction acting up the incline.
3. If the coefficient of static friction were 0.25 instead of 0.15, would the cylinder roll without slipping? Find its acceleration in that case.
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The required static-friction coefficient for rolling without slipping was found to be μ_req = 0.192. With μ_s = 0.25, the available static friction exceeds the required value, so rolling without slipping is possible.
The no-slip acceleration does not depend on the actual value of static friction as long as the limit is not exceeded: a = g sinθ/(1 + I/(M R²)) = g sinθ/(1 + 0.5) = (9.8)(0.5)/1.5 = 3.27 m/s².
The required friction would again be f = 0.5 M a = 3.27 N, and N = 16.97 N, giving μ_req = 0.192 < 0.25. The cylinder rolls without slipping with a = 3.27 m/s².