Question

# An 6.90-cm-diameter, 360 g solid sphere is released from rest at the top of a 1.80-m-long,...

An 6.90-cm-diameter, 360 g solid sphere is released from rest at the top of a 1.80-m-long, 20.0 ∘ incline. It rolls, without slipping, to the bottom. What is the sphere's angular velocity at the bottom of the incline? What fraction of its kinetic energy is rotational?

mass of the sphere, m=360 g

diameter, d=6.9cm

distance, d=1.8 m

incline angle, theta=20 degrees

a)

by using law of conservation energy,

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

m*g*h=7/10*m*v^2

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

here,

h=d*sin(theta)

V=sqrt(10*g*d*sin(theta)/7)

R*w=sqrt(10*g*d*sin(theta)/7)

3.45*10^-2*w=sqrt(10*9.8*1.8*sin(20)/7)

b)

rotational K.E=1/2*I*w^2

=1/2*(2/5*m*R^2)*w^2

=1/5**m*(R*w)^2

=1/5*m*v^2

and

Total K.E=7/10*m*v^2

===>

rotational K.E/total K.E=((1/5)*m*v^2)/((7/10)*m*v^2)

=(1/5)/(7/10)

=2/7

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