A solid sphere of uniform density starts from rest and rolls without slipping a

Question

A solid sphere of uniform density starts from rest and rolls without slipping a distance of d = 3 m down a q = 37° incline. The sphere has a mass M = 3.7 kg and a radius R = 0.28 m.1 Of the total kinetic energy of the sphere, what fraction is translational? 5/7

   OKb) What is the translational kinetic energy of the sphere when it reaches the bottom of the incline?

KEtran=

What is the translational speed of the sphere as it reaches the bottom of the ramp? v =

Now let's change the problem a little.

d) Suppose now that there is no frictional force between the sphere and the incline. Now, what is the translational kinetic energy of the sphere at the bottom of the incline?

KEtran

Explanation / Answer

a) using law of conservation of energy

m*g*h = 0.5*m*v^2 + 0.5*I*w^2

I is the moment of inertia = (2/5)*r*R^2

w = v/R

and h = l*sin(theta)

then

m*g*L*sin(theta) = KE_trans+ (0.5*(2/5)*m*R^2)*(v/R)^2 = (0.5*m*v^2) + (1/5)*m*v^2 = 0.7*m*v^2

required fraction is 0.5*m*v^2/ (0.7*m*v^2) = 5/7

b) KE_trans = (5/7)*m*g*l*sin(theta) = (5/7)*3.7*9.81*3*sin(37) = 46.8 J

c) 0.7*m*v^2 = m*g*l*sin(theta)

0.7*v^2 = 9.8*3*sin(37)

v^2 = (9.8*3*sin(37))/0.7 = 25.27

v = sqrt(25.27) = 5.02 m/sec

d) KE = 0.5*m*v^2 = m*g*l*sin(theta) = 3.7*9.81*3*sin(37) = 65.53 J

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