Question

A thin hollow cylinder
(a) slides without rotating with a speed v. (b) rolls with the same speed without slipping.
Find the ratio of kinetic energies in the two cases.​

Answer 1

Answer:

1:2

Explanation:

$\bold{As \: we \: know \: that \: for} \\ \bold{hollow \: cylinder \: \frac{ {K}^{2} }{ {R}^{2} } \: will \: be \: 1}.$

[ Reason- Analogous to a as ring ]

Now, we need to find the kinetic energy of both so,

$\bold{(a).} \: E_{trans} = \frac{1}{2}M {v}^{2} \\ \bold{(b).} \: E_{rolling} = \frac{1}{2}M {v}^{2}(1 + \frac{ {k}^{2} }{ {r}^{2} } ) \\ = > E_{rolling} = \frac{1}{2}M {v}^{2}(1 + 1) \\ = > E_{rolling} = \frac{1}{2}M {v}^{2}(2) \\ = > E_{rolling} = \frac{ \cancel2M {v}^{2} }{ \cancel2} \\ = > E_{rolling} = M {v}^{2}$

Now finally the required ratio-:

$= > \frac{E_{trans}}{E_{rolling}} = \frac{ \frac{1}{2}M {v}^{2} }{M {v}^{2} } \\ = > \frac{E_{trans}}{E_{rolling}} = \frac{ \frac{M {v}^{2} }{2} }{M {v}^{2} } \\ = > \frac{E_{trans}}{E_{rolling}} = \frac{\cancel{M} \cancel{{v}}^{2} }{2} \times \frac{1}{ \cancel{M} { \cancel{v}}^{2} } \\ = > \boxed{\frac{E_{trans}}{E_{rolling}} = \frac{1}{2}}$

Therefore-:

The ratio of kinetic energies will be 1:2

Answer 2

Answer:

1:2

Explanation:

Solution is in the attachment