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 = 4.2 m down a ? = 25°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?

KE tran/KEtotal =

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

KE tran =

3) What is the translational speed of the sphere as it reaches the bottom of the ramp?

v =

4) Now let's change the problem a little.

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?

KE tran =

Just need help understanding how to attack these questions. Thanks!

Explanation / Answer

A.

Using conservation of energy,

initial PE + KE = final PE + KE

m g h + 0 = 0 + (m v^2 /2 + I w^2 / 2)

m g d sin25 = m v^2 / 2 + (2 m r^2 / 5) ( v/r)^2 / 2

g d sin25 = v^2 / 2 + v^2 / 5 = 0.7 v^2

v^2 = 9.8 x 4.2 x sin25/0. 7

v^= 24.85

v = 4.985 m/s

total KE = 0.7 m v^2= 64.362 J

translational KE = m v^2 /2 = 3.7 x 4.985^2 / 2 = 45.973 J

fraction = 45.973 / 64.362 =0.714

------------------

(2) trans KE = 45.973 J

------------------

(3) v = 4.985 m/s

-----------------
(4) when there is no friction then

rotational Ke = 0

trans KE = total energy = m g d sin25 = 64.36 J

----------------

Comment in case any doubt..goodluck

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.