A solid sphere (radius R = 5.0 cm and M = 100.0 g) is released from rest at the

Question

A solid sphere (radius R = 5.0 cm and M = 100.0 g) is released from rest at the top of an inclined plane with height 3.0 m and inclination of 45 degree rolling down without slipping. Calculate (i) the speed at the center of mass at the bottom, (ii) the magnitude of linear acceleration at the center of sphere, (iii) the time required for the sphere to reach the bottom, and (iv) the magnitude of the friction force on the sphere. Moment of inertia of a sphere is 2MR^2/5, where m is the mass of the sphere.

Explanation / Answer

i) Apply conservation of energy

final kinetic energy = initial potential energy

(1/2)*M*v^2 + (1/2)*I*w^2 = M*g*h

(1/2)*M*v^2 + (1/2)*(2/5)*M*R^2*w^2 = M*g*h

(1/2)*M*v^2 + (1/5)*M*(R*w)^2 = M*g*h

(1/2)*M*v^2 + (1/5)*M*v^2 = M*g*h

(7/10)*M*v^2 = M*g*h

==> v = sqrt(g*h*10/7)

= sqrt(9.8*3*10/7)

= 6.48 m/s

ii) distance travelled along the incline,

d = h/sin(45)

= 3/sin(45)

= 4.24 m

now use, v^2 - u^2 = 2*a*d

v^2 - 0 = 2*a*d

==> a = v^2/(2*d)

= 6.48^2/(2*4.24)

= 4.95 m/s^2

iii) use, v = u + a*t

==> t = (v - u)/a

= (6.48 - 0 )/4.95

= 1.31 s

iv) net force, Fnet = m*g*sin(45) - fs

M*a = M*g*sin(45) - fs

==> fs = M*g*sin(45) - M*a

= 0.1*9.8*sin(45) - 0.1*4.95

= 0.198 N

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