Question

Q2. A sphere of radius r and weight W is released with no initial velocity on a (C*) degrees incline and rolls without slipping. Determine: (a) the minimum value of the coefficient of static friction (µ) compatible with the rolling motion, (b) the velocity (Vb) of the center G of the sphere after the sphere has rolled (D*) meters, (c) the velocity (Vc) of G if the sphere were to move (D*) meters down a frictionless (C*) degrees incline. (g = 9.81m/s2) Report the following in the results summary table: part (a): µ = ? (10 points) part (b) Vb = ? (m/s) (10 points) part (c) Vc =? (m/s) (10 points)

C=42.3

D=23.5

Answer 1

Ans a) The external forces W, N and F form a system equivalent to the system of effective forces represented by vector m and couple .Since, sphere rolls without sliding we have = r

=> (W Sin ) r = (m)r +

=> (W Sin ) r = (m r ) r +

where, W = mg and   = (2/5) m

=> (mg Sin )r = m + (2/5) m

=> g Sin = r + (2/5) r

=> g Sin = (7/5) r

=> = 5 g Sin / 7 r

Also, = r

=> = 5 g Sin / 7

=> = 5(9.81) Sin(42.3) / 7

=> = 4.716 m/

Equating forces in X direction, Fx = m

=> mg Sin(42.3) - F = m(4.716)

=> m(9.81)(0.673) - F = 4.716 m

=> F = 1.886 m .........................(1)

Equating forces in vertical direction,  Fy = 0

=> N - mg Cos(42.3) = 0

=> N = m(9.81)(0.74)

=> N = 7.255 m.......................(2)

Hence, coefficient of static friction () = F / N = 1.886 m / 7.255 m

=> = 0.26

.

Ans b) We know, for uniformly accelerated motion,

= + 2 D

Initial velocity , u = 0

=> = 0 + 2(4.716)(23.5)

=> = 221.65

=> = 14.88 m/s

.

Ans c) Now, since there is no friction , we have F = 0,

=>   = 0

=> = 0

Equating forces in X direction, Fx = m

=> mg Sin(42.3) = m

=> = 9.81(0.673) = 6.60 m/

Hence, putting values in equation of uniform motion,

= + 2 D

Initial velocity , u = 0

=> = 0 + 2(6.60)(23.5)

=> = 310.2

=> = 17.61 m/s