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?

Homework Answers

Answer #1


mass of the sphere, m=360 g


diameter, d=6.9cm


radius, R=d/2=6.9/2=3.45cm


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)


===> w=85.1 rad/sec


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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