Asked by Dan
An 9.00-cm-diameter, 400 sphere is released from rest at the top of a 1.90-m-long, 20.0 incline. It rolls, without slipping, to the bottom.
A) What is the sphere's angular velocity at the bottom of the incline?
B) WWhat fraction of its kinetic energy is rotational?
A) What is the sphere's angular velocity at the bottom of the incline?
B) WWhat fraction of its kinetic energy is rotational?
Answers
Answered by
bobpursley
mgh=rotational energy + translational energy
I assume the 20 is an angle
mg(1.90 sin20)=1/2 I w^2+1/2 mv^2
but v=wr so
= 1/2 I w^2+1/2 m w^2 r^2
so calculate w.
then you have the two terms, and the total energy, solve for the fraction.
I assume the 20 is an angle
mg(1.90 sin20)=1/2 I w^2+1/2 mv^2
but v=wr so
= 1/2 I w^2+1/2 m w^2 r^2
so calculate w.
then you have the two terms, and the total energy, solve for the fraction.
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