Three bodies a ring a solid cylinder and a solid sphere roll down
Answers
Answer:
We assume conservation of energy of the rolloing body, i.e. there is no loss of energy due to friction etc. The potential energy lost by the body in rolling down the inclined plane (=mgh)(=mgh) must, therefore, be equal to kinetic energy gained. Since the bodies start from rest the kinetic energy gained is equal to the final kinetic energy of the bodies. From Eq., K=\displaystyle\frac{1}{2}mv^2\left(1+\frac{k^2}{R^2}\right)K=
2
1
mv
2
(1+
R
2
k
2
), where vv is the final velocity of (the centre of mass of) the body. Equating KK and mgh,mgh,
mgh=\displaystyle\frac{1}{2}mv^2\left(1+\displaystyle\frac{k^2}{R^2}\right)mgh=
2
1
mv
2
(1+
R
2
k
2
)
or v^2=\left(\displaystyle\frac{2gh}{1+k^2/R^2}\right)v
2
=(
1+k
2
/R
2
2gh
)
Note is independent of the mass of the rolling body;;
For a ring, k^2=R^2k
2
=R
2
v_{ring}=\displaystyle\sqrt{\frac{2gh}{1+1}},v
ring
=
1+1
2gh
,
=\sqrt{gh}=
gh
For a solid cylinder k^2=R^2/2k
2
=R
2
/2
v_{disc}=\sqrt{\displaystyle\frac{2gh}{1+1/2}}v
disc
=
1+1/2
2gh
=\sqrt{\displaystyle\frac{4gh}{3}}=
3
4gh
For a solid sphere k^2=2R^2/5k
2
=2R
2
/5
v_{sphere}=\sqrt{\displaystyle\frac{2gh}{1+2/5}}v
sphere
=
1+2/5
2gh
=\sqrt{\displaystyle\frac{10gh}{7}}=
7
10gh
From the results obtained it is clear that among the three bodies the sphere has the greatest and the ring has the least velocity of the centre of mass at the bottom of the inclined plane.