A hollow, spherical shell with mass 2.45 kg rolls without slipping down a slope

Question

A hollow, spherical shell with mass 2.45 kg rolls without slipping down a slope angled at 36.0 degrees

Part A

find the acceleration
-take free fall acceleration to be g= 9.8 m/s squared

Part B
find the friction force
-take the free fall accelertion to be g=9.8 m/s^2

Part C
Find the minimum coefficient of friction needed to prevent slipping

Explanation / Answer

(A) a = gsin36 = 9.8 sin 36 = 5.76m/s2 Ans; (B) Friction force x radius = torque = moment of inertia x radial acceleration, so F x r = 0.67mr^2 x a/r --> F = 0.67ma = 0.67 x 2.45 x 5.76 = 9.408 N Ans.; (C) coefficient of friction x normal force = force of friction ---> coefficient of friction x mgcos36 = 9.408N, so coefficient of friction = 9.408/(2.45 x 9.8cos36) = 9.408 / 19.472 = 0.4843 Ans.

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