1) Of the total kinetic energy of the sphere, what fraction is translational? KE

ID: 1286696 • Letter: 1

Question

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 =

theta = 35 incline. The sphere has a mass M = 3.1 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 = A solid sphere of uniform density starts from rest and rolls without slipping a distance of d = 4.7 m down a

Explanation / Answer

a) KE at any point = Translational KE + Rotational KE
= ( 0.5 . m . v^2 ) + (0.5 . I . w^2 )

I = (2/5) . m . r^2
w = v/r

so KE = 0.5 . m . v^2 + 0.5 . 0.4 . m . r^2 . v^2/r^2
KE = 0.7 .m . v^2

So the translational KE is 5/7 ths of the total KE.

b) At the bottom all the height energy lost has been converted into KE.
Of that KE 5/7 ths is translational KE.

Height energy lost = m . g. h = m .g . 4.7 . sin35?

Translational KE = 5/7 . 3.1 . 9.8 . 4.7 . sin(35) = 58.5 J

c) Just equate the 58.5 J with 0.5 . m . v^2 and solve for v. It comes to 6.14 m/s

d) If there is no friction the ball will slide without rolling so there is no rotational KE to take into account.

So m.g.h = 0.5 . m . v^2

v = ?( 2 . 9.8 . 4.7 . sin 35) = 7.27 m/s

So KEtran = 0.5*3.1*7.27^2 = 81.9 J

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1) Of the total kinetic energy of the sphere, what fraction is translational? KE

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