Question

A solid sphere ( of mass 2.50 kg and radius 10.0 cm) starts rolling without slipping on an inclined plane (angle of inclination 30 deg). Find the speed of its center of mass when it has traveled down 2.00 m along with the inclination.

Groups of choices:

a. 3.13 m/s

b. 4.43 m/s

c. 3.74 m/s

d. 6.26 m/s

Answer 1

Given Solid sphere of mass m = 2.5 kg, radius R=10.0 cm

on an 30 degrees inclined plane,rolling without slipping
traveled a distance of 2 m along the incline

from the data we can calculate the height of the incline h,

sin 30 = h/2 ==> h = 2 sin 30 =2*0.5 = 1 m

assuming that there is no friction then by conservation of energy

initially the total energy is gravitational potential energy and later kinetic energy
here the sphere will have both rotational kinetic energy and translational kinetic energy
so

mgh = 0.5*I*W^2 +0.5*m*v^2
mgh = 0.5*(2/5)*m*R^2*W^2 + 0.5*m*v^2

mgh = 0.5*(2/5)*m*R^2*V^2/R^2+0.5*m*v^2

gh = 0.5*(2/5)*V^2+0.5*v^2

gh = 0.5*v^2(2/5 +1)

v^2 = 2gh/(2/5 +1)
v = sqrt(2gh/(2/5 +1))

V = sqrt(2gh/(2/5 +1))

substituting the values

V = sqrt((2*9.8*1)/(2/5+1)) m/s

V = 3.74 m/s