Question

A solid sphere and a hoop start from rest and roll down and incline plane (without...

A solid sphere and a hoop start from rest and roll down and incline plane (without slipping). Assume they both have the same mass and radius.

(a) Draw the free body diagram and calculate the acceleration of each.

(b) What is the distance between them after 4.00 s and which one is in front?

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Answer #1

(a) First for sphere
We know that
torque = I
where is angular acceleration = a/r
wherer a is linear acceleration and r is radius
I is moment of inertia = (2/5)mr2
m is mass of sphere
Now friction force wil provide torque , therefore
umgCos*r = (2/5)mr2*(as/r) ----------(1)
Using the second law we can write
mgSin - umgCos = mas ----------(2)
On solving 1 and 2 , we get
as = (5/7)gSin
Now for hoop
Moment of inertia (i) = mr2
therefore
umgCos*r = (mr2*(ah/r) ---------(3)
mgSin - umgCos​ = mah ----------(4)
Now solving 3 and 4, we get
ah = (1/2)gSin
(b)
Now the distance calculated by the sphere
SS = ut +(1/2) ast2
since initially it is at rest , therefore u = 0
Ss = 0 +(1/2)*(5/7)gSin*(4)2 = 56.057Sin
Now by the hoop
Sh = (1/2)ah*(t2 )= (1/2)*(1/2)*9.81*Sin*(4)2= 39.24Sin
Hence sphere will be ahead then hoop.

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