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Rotational Motion

150404 A solid cylinder is released from rest from the top of an inclined plane of inclination $30^{\circ}$ and length $60 cm$ . If the cylinder rolls without slipping, then the speed when it reaches the bottom is

1

1.5 m / s

2

2.0 m / s

3

3.0 m / s

4

6.0 m / s

Explanation:

B Given,
$θ = 30^{\circ}, l = 60 cm = 60 \times 10^{- 2} m$
By work energy theorem,
$mg (l \sin θ) = \frac{1}{2} I ω^{2} + \frac{1}{2} {mv}^{2}$
$mg (l \sin θ) = \frac{1}{2} (\frac{1}{2} {mr}^{2}) {(\frac{v}{r})}^{2} + \frac{1}{2} {mv}^{2} (∵ v = r ω$
$mg (l \sin θ) = (\frac{1}{4} + \frac{1}{2}) {mv}^{2}$
$g l \sin θ = (\frac{1 + 2}{4}) v^{2}$
$g l \sin θ = \frac{3}{4} v^{2}$
$v^{2} = \frac{4 g l \sin θ}{3}$
$v = \sqrt{\frac{4 g l \sin θ}{3}}$
$v = \sqrt{\frac{4 \times 9.8 \times 60 \times 10^{- 2} \times \sin 30^{\circ}}{3}}$
$v = 1.92$
$v ≃ 2 m / s$

TS- EAMCET.11.09.2020

Rotational Motion

150405 A solid sphere of mass $2 kg$ rolls on a smooth horizontal surface at $10 m / s$ . It then rolls up a smooth inclined plane of inclination $30^{\circ}$ with the horizontal. The height attained by the sphere before it stops is [take $g = 10 m / s^{2}$ ]

1

70 cm

2

701 cm

3

7.0 m

4

70 m

Explanation:

C Given,
Mass of solid sphere $= 2 kg$
Velocity $v) = 10 m / s$
Angle $θ) = 30^{\circ}$
original image
If sphere is at point $P$ , sphere has kinetic energy and rotational energy and when it is at point $Q$ , it have only potential energy.
From law of conservation of energy,
$\frac{1}{2} {mv}^{2} + \frac{1}{2} I ω^{2} = mgh$
$\frac{1}{2} {mv}^{2} + \frac{1}{2} \times \frac{2}{5} {mr}^{2} \times \frac{v^{2}}{r^{2}} = mgh [\begin{array}{l} ∴ I = \frac{2}{5} {mr}^{2} \\ and ω = \frac{v}{r} \end{array}]$
${mv}^{2} [\frac{1}{2} + \frac{1}{5}] = mgh$
$\frac{7 v^{2}}{10} = gh$
Put the value of $v$ and $g$ ,
$h = \frac{7 \times (10)^{2}}{10 \times 10}$
$h = 7 m$

TS- EAMCET-11.09.2020

Rotational Motion

150406 A solid sphere and a solid cylinder, each of mass $M$ and radius $R$ are rolling with a linear speed on a flat surface without slipping. Let $L_{1}$ be magnitude of the angular momentum of the sphere with respect to a fixed point along the path of the sphere. Likewise $L_{2}$ be the magnitude of angular momentum of the cylinder with respect to the same fixed point along its path. The ratio $\frac{L_{1}}{L_{2}}$ is

1

\frac{14}{15}

2

\frac{4}{5}

3

\frac{2}{5}

4

\frac{7}{15}

Explanation:

A Given,
Mass of solid sphere and cylinder $= M$
Radius of solid sphere and cylinder $= R$
$L_{1}$ and $L_{2}$ are the magnitude of the angular momentum of the sphere and cylinder.

For linear motion $(L_{1}) = MvR$
$= M (R ω) \times R (∴ v = R ω$
$= {MR}^{2} \times ω$
For rotational motion $(L_{2}) = I ω$
So,
Angular momentum about origin ' $O$ ' $(L_{o}) = L_{1} + L_{2}$
$= {MR}^{2} ω + I ω$
$= ω ({MR}^{2} + I$
We know,
Moment of inertia for solid sphere, $I = \frac{2}{5} {MR}^{2}$
And for solid cylinder, $I = \frac{1}{2} {MR}^{2}$
So, ${(L_{1})}_{about 'O'} = ({MR}^{2} + \frac{2}{5} {MR}^{2}) ω = \frac{7}{5} {MR}^{2} ω$
And ${(L_{2})}_{about ' O^{'}} = ({MR}^{2} + \frac{1}{2} {MR}^{2}) ω = \frac{3}{2} {MR}^{2} ω$
So,
$\frac{L_{1}}{L_{2}} = \frac{\frac{7}{5} {MR}^{2} ω}{\frac{3}{2} {MR}^{2} ω}$
$\frac{L_{1}}{L_{2}} = \frac{14}{15}$

TS- EAMCET-10.09.2020

Rotational Motion

150407 A solid sphere of mass $2 kg$ rolls on a smooth horizontal surface at $10 m / s$ and then rolls up a smooth inclined plane of inclination $30^{\circ}$ with the horizontal. The maximum height reached by the sphere is $(g = 9.8 m / s^{2})$

1

10 m

2

4.9 m

3

14.2 m

4

7.1 m

Explanation:

D Given -
Mass of solid sphere $m) = 2 kg$
Velocity $v) = 10 m / s$

When the sphere is at point $A$ , it possesses kinetic energy and rotational energy and when it is at point $B$ it possesses only potential energy.
So from the law of conservation of energy -
$\frac{1}{2} {mv}^{2} + \frac{1}{2} I ω^{2} = mgh$
$Or \frac{1}{2} {mv}^{2} + \frac{1}{2} {mk}^{2} \times \frac{v^{2}}{R^{2}} = mgh$
$Or \frac{1}{2} {mv}^{2} [1 + \frac{k^{2}}{R^{2}}] = mgh$
$Or$
$\frac{1}{2} \times (10)^{2} [1 + \frac{2}{5}] = 9.8 \times h {$ For solid sphere, $\frac{k^{2}}{R^{2}} = \frac{2}{5}}$
$\frac{1}{2} \times 100 \times \frac{7}{5} = 9.8 h [\begin{array}{l} 70 = 9.8 h \\ h = \frac{700}{98} = 7.1 \end{array}]$
$h = 7.1 m$

Manipal UGET -2020

Rotational Motion

150408 A solid cylinder and a hollow cylinder both of the same mass and same external diameter are released from the same height at the same time on an inclined plane. Both roll down without slipping. Which one will reach the bottom first?

1 Solid cylinder

2 Both together

3 One with higher density

4 Hollow cylinder

Explanation:

A For a rolling body, acceleration is given by
$a = \frac{g \sin θ}{(1 + \frac{k^{2}}{R^{2}})}$
For solid sphere $I = \frac{2}{5} {MR}^{2}$
If radius of gyration is same for both cylinder then
${Mk}^{2} = \frac{2}{5} {MR}^{2}$
$\frac{k^{2}}{R^{2}} = \frac{2}{5}$
$a_{solid} = \frac{5}{7} g \sin θ = 0.71 g \sin θ$
For hollow sphere $I = \frac{1}{2} {MR}^{2}$
${Mk}^{2} = \frac{1}{2} {MR}^{2} \Rightarrow \frac{k^{2}}{R^{2}} = \frac{1}{2}$
$a_{hollow} = \frac{2}{3} g \sin θ = 0.66 g \sin θ$
So,
$a_{solid} > a_{hollow}$
Hence, solid will reach the bottom first.

BITSAT-2020

Rotational Motion

150404 A solid cylinder is released from rest from the top of an inclined plane of inclination $30^{\circ}$ and length $60 cm$ . If the cylinder rolls without slipping, then the speed when it reaches the bottom is

1

1.5 m / s

2

2.0 m / s

3

3.0 m / s

4

6.0 m / s

Explanation:

B Given,
$θ = 30^{\circ}, l = 60 cm = 60 \times 10^{- 2} m$
By work energy theorem,
$mg (l \sin θ) = \frac{1}{2} I ω^{2} + \frac{1}{2} {mv}^{2}$
$mg (l \sin θ) = \frac{1}{2} (\frac{1}{2} {mr}^{2}) {(\frac{v}{r})}^{2} + \frac{1}{2} {mv}^{2} (∵ v = r ω$
$mg (l \sin θ) = (\frac{1}{4} + \frac{1}{2}) {mv}^{2}$
$g l \sin θ = (\frac{1 + 2}{4}) v^{2}$
$g l \sin θ = \frac{3}{4} v^{2}$
$v^{2} = \frac{4 g l \sin θ}{3}$
$v = \sqrt{\frac{4 g l \sin θ}{3}}$
$v = \sqrt{\frac{4 \times 9.8 \times 60 \times 10^{- 2} \times \sin 30^{\circ}}{3}}$
$v = 1.92$
$v ≃ 2 m / s$

TS- EAMCET.11.09.2020

Rotational Motion

150405 A solid sphere of mass $2 kg$ rolls on a smooth horizontal surface at $10 m / s$ . It then rolls up a smooth inclined plane of inclination $30^{\circ}$ with the horizontal. The height attained by the sphere before it stops is [take $g = 10 m / s^{2}$ ]

1

70 cm

2

701 cm

3

7.0 m

4

70 m

Explanation:

C Given,
Mass of solid sphere $= 2 kg$
Velocity $v) = 10 m / s$
Angle $θ) = 30^{\circ}$
original image
If sphere is at point $P$ , sphere has kinetic energy and rotational energy and when it is at point $Q$ , it have only potential energy.
From law of conservation of energy,
$\frac{1}{2} {mv}^{2} + \frac{1}{2} I ω^{2} = mgh$
$\frac{1}{2} {mv}^{2} + \frac{1}{2} \times \frac{2}{5} {mr}^{2} \times \frac{v^{2}}{r^{2}} = mgh [\begin{array}{l} ∴ I = \frac{2}{5} {mr}^{2} \\ and ω = \frac{v}{r} \end{array}]$
${mv}^{2} [\frac{1}{2} + \frac{1}{5}] = mgh$
$\frac{7 v^{2}}{10} = gh$
Put the value of $v$ and $g$ ,
$h = \frac{7 \times (10)^{2}}{10 \times 10}$
$h = 7 m$

TS- EAMCET-11.09.2020

Rotational Motion

150406 A solid sphere and a solid cylinder, each of mass $M$ and radius $R$ are rolling with a linear speed on a flat surface without slipping. Let $L_{1}$ be magnitude of the angular momentum of the sphere with respect to a fixed point along the path of the sphere. Likewise $L_{2}$ be the magnitude of angular momentum of the cylinder with respect to the same fixed point along its path. The ratio $\frac{L_{1}}{L_{2}}$ is

1

\frac{14}{15}

2

\frac{4}{5}

3

\frac{2}{5}

4

\frac{7}{15}

Explanation:

A Given,
Mass of solid sphere and cylinder $= M$
Radius of solid sphere and cylinder $= R$
$L_{1}$ and $L_{2}$ are the magnitude of the angular momentum of the sphere and cylinder.

For linear motion $(L_{1}) = MvR$
$= M (R ω) \times R (∴ v = R ω$
$= {MR}^{2} \times ω$
For rotational motion $(L_{2}) = I ω$
So,
Angular momentum about origin ' $O$ ' $(L_{o}) = L_{1} + L_{2}$
$= {MR}^{2} ω + I ω$
$= ω ({MR}^{2} + I$
We know,
Moment of inertia for solid sphere, $I = \frac{2}{5} {MR}^{2}$
And for solid cylinder, $I = \frac{1}{2} {MR}^{2}$
So, ${(L_{1})}_{about 'O'} = ({MR}^{2} + \frac{2}{5} {MR}^{2}) ω = \frac{7}{5} {MR}^{2} ω$
And ${(L_{2})}_{about ' O^{'}} = ({MR}^{2} + \frac{1}{2} {MR}^{2}) ω = \frac{3}{2} {MR}^{2} ω$
So,
$\frac{L_{1}}{L_{2}} = \frac{\frac{7}{5} {MR}^{2} ω}{\frac{3}{2} {MR}^{2} ω}$
$\frac{L_{1}}{L_{2}} = \frac{14}{15}$

TS- EAMCET-10.09.2020

Rotational Motion

150407 A solid sphere of mass $2 kg$ rolls on a smooth horizontal surface at $10 m / s$ and then rolls up a smooth inclined plane of inclination $30^{\circ}$ with the horizontal. The maximum height reached by the sphere is $(g = 9.8 m / s^{2})$

1

10 m

2

4.9 m

3

14.2 m

4

7.1 m

Explanation:

D Given -
Mass of solid sphere $m) = 2 kg$
Velocity $v) = 10 m / s$

When the sphere is at point $A$ , it possesses kinetic energy and rotational energy and when it is at point $B$ it possesses only potential energy.
So from the law of conservation of energy -
$\frac{1}{2} {mv}^{2} + \frac{1}{2} I ω^{2} = mgh$
$Or \frac{1}{2} {mv}^{2} + \frac{1}{2} {mk}^{2} \times \frac{v^{2}}{R^{2}} = mgh$
$Or \frac{1}{2} {mv}^{2} [1 + \frac{k^{2}}{R^{2}}] = mgh$
$Or$
$\frac{1}{2} \times (10)^{2} [1 + \frac{2}{5}] = 9.8 \times h {$ For solid sphere, $\frac{k^{2}}{R^{2}} = \frac{2}{5}}$
$\frac{1}{2} \times 100 \times \frac{7}{5} = 9.8 h [\begin{array}{l} 70 = 9.8 h \\ h = \frac{700}{98} = 7.1 \end{array}]$
$h = 7.1 m$

Manipal UGET -2020

Rotational Motion

150408 A solid cylinder and a hollow cylinder both of the same mass and same external diameter are released from the same height at the same time on an inclined plane. Both roll down without slipping. Which one will reach the bottom first?

1 Solid cylinder

2 Both together

3 One with higher density

4 Hollow cylinder

Explanation:

A For a rolling body, acceleration is given by
$a = \frac{g \sin θ}{(1 + \frac{k^{2}}{R^{2}})}$
For solid sphere $I = \frac{2}{5} {MR}^{2}$
If radius of gyration is same for both cylinder then
${Mk}^{2} = \frac{2}{5} {MR}^{2}$
$\frac{k^{2}}{R^{2}} = \frac{2}{5}$
$a_{solid} = \frac{5}{7} g \sin θ = 0.71 g \sin θ$
For hollow sphere $I = \frac{1}{2} {MR}^{2}$
${Mk}^{2} = \frac{1}{2} {MR}^{2} \Rightarrow \frac{k^{2}}{R^{2}} = \frac{1}{2}$
$a_{hollow} = \frac{2}{3} g \sin θ = 0.66 g \sin θ$
So,
$a_{solid} > a_{hollow}$
Hence, solid will reach the bottom first.

BITSAT-2020

Rotational Motion

150404 A solid cylinder is released from rest from the top of an inclined plane of inclination $30^{\circ}$ and length $60 cm$ . If the cylinder rolls without slipping, then the speed when it reaches the bottom is

1

1.5 m / s

2

2.0 m / s

3

3.0 m / s

4

6.0 m / s

Explanation:

B Given,
$θ = 30^{\circ}, l = 60 cm = 60 \times 10^{- 2} m$
By work energy theorem,
$mg (l \sin θ) = \frac{1}{2} I ω^{2} + \frac{1}{2} {mv}^{2}$
$mg (l \sin θ) = \frac{1}{2} (\frac{1}{2} {mr}^{2}) {(\frac{v}{r})}^{2} + \frac{1}{2} {mv}^{2} (∵ v = r ω$
$mg (l \sin θ) = (\frac{1}{4} + \frac{1}{2}) {mv}^{2}$
$g l \sin θ = (\frac{1 + 2}{4}) v^{2}$
$g l \sin θ = \frac{3}{4} v^{2}$
$v^{2} = \frac{4 g l \sin θ}{3}$
$v = \sqrt{\frac{4 g l \sin θ}{3}}$
$v = \sqrt{\frac{4 \times 9.8 \times 60 \times 10^{- 2} \times \sin 30^{\circ}}{3}}$
$v = 1.92$
$v ≃ 2 m / s$

TS- EAMCET.11.09.2020

Rotational Motion

150405 A solid sphere of mass $2 kg$ rolls on a smooth horizontal surface at $10 m / s$ . It then rolls up a smooth inclined plane of inclination $30^{\circ}$ with the horizontal. The height attained by the sphere before it stops is [take $g = 10 m / s^{2}$ ]

1

70 cm

2

701 cm

3

7.0 m

4

70 m

Explanation:

C Given,
Mass of solid sphere $= 2 kg$
Velocity $v) = 10 m / s$
Angle $θ) = 30^{\circ}$
original image
If sphere is at point $P$ , sphere has kinetic energy and rotational energy and when it is at point $Q$ , it have only potential energy.
From law of conservation of energy,
$\frac{1}{2} {mv}^{2} + \frac{1}{2} I ω^{2} = mgh$
$\frac{1}{2} {mv}^{2} + \frac{1}{2} \times \frac{2}{5} {mr}^{2} \times \frac{v^{2}}{r^{2}} = mgh [\begin{array}{l} ∴ I = \frac{2}{5} {mr}^{2} \\ and ω = \frac{v}{r} \end{array}]$
${mv}^{2} [\frac{1}{2} + \frac{1}{5}] = mgh$
$\frac{7 v^{2}}{10} = gh$
Put the value of $v$ and $g$ ,
$h = \frac{7 \times (10)^{2}}{10 \times 10}$
$h = 7 m$

TS- EAMCET-11.09.2020

Rotational Motion

150406 A solid sphere and a solid cylinder, each of mass $M$ and radius $R$ are rolling with a linear speed on a flat surface without slipping. Let $L_{1}$ be magnitude of the angular momentum of the sphere with respect to a fixed point along the path of the sphere. Likewise $L_{2}$ be the magnitude of angular momentum of the cylinder with respect to the same fixed point along its path. The ratio $\frac{L_{1}}{L_{2}}$ is

1

\frac{14}{15}

2

\frac{4}{5}

3

\frac{2}{5}

4

\frac{7}{15}

Explanation:

A Given,
Mass of solid sphere and cylinder $= M$
Radius of solid sphere and cylinder $= R$
$L_{1}$ and $L_{2}$ are the magnitude of the angular momentum of the sphere and cylinder.

For linear motion $(L_{1}) = MvR$
$= M (R ω) \times R (∴ v = R ω$
$= {MR}^{2} \times ω$
For rotational motion $(L_{2}) = I ω$
So,
Angular momentum about origin ' $O$ ' $(L_{o}) = L_{1} + L_{2}$
$= {MR}^{2} ω + I ω$
$= ω ({MR}^{2} + I$
We know,
Moment of inertia for solid sphere, $I = \frac{2}{5} {MR}^{2}$
And for solid cylinder, $I = \frac{1}{2} {MR}^{2}$
So, ${(L_{1})}_{about 'O'} = ({MR}^{2} + \frac{2}{5} {MR}^{2}) ω = \frac{7}{5} {MR}^{2} ω$
And ${(L_{2})}_{about ' O^{'}} = ({MR}^{2} + \frac{1}{2} {MR}^{2}) ω = \frac{3}{2} {MR}^{2} ω$
So,
$\frac{L_{1}}{L_{2}} = \frac{\frac{7}{5} {MR}^{2} ω}{\frac{3}{2} {MR}^{2} ω}$
$\frac{L_{1}}{L_{2}} = \frac{14}{15}$

TS- EAMCET-10.09.2020

Rotational Motion

150407 A solid sphere of mass $2 kg$ rolls on a smooth horizontal surface at $10 m / s$ and then rolls up a smooth inclined plane of inclination $30^{\circ}$ with the horizontal. The maximum height reached by the sphere is $(g = 9.8 m / s^{2})$

1

10 m

2

4.9 m

3

14.2 m

4

7.1 m

Explanation:

D Given -
Mass of solid sphere $m) = 2 kg$
Velocity $v) = 10 m / s$

When the sphere is at point $A$ , it possesses kinetic energy and rotational energy and when it is at point $B$ it possesses only potential energy.
So from the law of conservation of energy -
$\frac{1}{2} {mv}^{2} + \frac{1}{2} I ω^{2} = mgh$
$Or \frac{1}{2} {mv}^{2} + \frac{1}{2} {mk}^{2} \times \frac{v^{2}}{R^{2}} = mgh$
$Or \frac{1}{2} {mv}^{2} [1 + \frac{k^{2}}{R^{2}}] = mgh$
$Or$
$\frac{1}{2} \times (10)^{2} [1 + \frac{2}{5}] = 9.8 \times h {$ For solid sphere, $\frac{k^{2}}{R^{2}} = \frac{2}{5}}$
$\frac{1}{2} \times 100 \times \frac{7}{5} = 9.8 h [\begin{array}{l} 70 = 9.8 h \\ h = \frac{700}{98} = 7.1 \end{array}]$
$h = 7.1 m$

Manipal UGET -2020

Rotational Motion

150408 A solid cylinder and a hollow cylinder both of the same mass and same external diameter are released from the same height at the same time on an inclined plane. Both roll down without slipping. Which one will reach the bottom first?

1 Solid cylinder

2 Both together

3 One with higher density

4 Hollow cylinder

Explanation:

A For a rolling body, acceleration is given by
$a = \frac{g \sin θ}{(1 + \frac{k^{2}}{R^{2}})}$
For solid sphere $I = \frac{2}{5} {MR}^{2}$
If radius of gyration is same for both cylinder then
${Mk}^{2} = \frac{2}{5} {MR}^{2}$
$\frac{k^{2}}{R^{2}} = \frac{2}{5}$
$a_{solid} = \frac{5}{7} g \sin θ = 0.71 g \sin θ$
For hollow sphere $I = \frac{1}{2} {MR}^{2}$
${Mk}^{2} = \frac{1}{2} {MR}^{2} \Rightarrow \frac{k^{2}}{R^{2}} = \frac{1}{2}$
$a_{hollow} = \frac{2}{3} g \sin θ = 0.66 g \sin θ$
So,
$a_{solid} > a_{hollow}$
Hence, solid will reach the bottom first.

BITSAT-2020

Rotational Motion

150404 A solid cylinder is released from rest from the top of an inclined plane of inclination $30^{\circ}$ and length $60 cm$ . If the cylinder rolls without slipping, then the speed when it reaches the bottom is

1

1.5 m / s

2

2.0 m / s

3

3.0 m / s

4

6.0 m / s

Explanation:

B Given,
$θ = 30^{\circ}, l = 60 cm = 60 \times 10^{- 2} m$
By work energy theorem,
$mg (l \sin θ) = \frac{1}{2} I ω^{2} + \frac{1}{2} {mv}^{2}$
$mg (l \sin θ) = \frac{1}{2} (\frac{1}{2} {mr}^{2}) {(\frac{v}{r})}^{2} + \frac{1}{2} {mv}^{2} (∵ v = r ω$
$mg (l \sin θ) = (\frac{1}{4} + \frac{1}{2}) {mv}^{2}$
$g l \sin θ = (\frac{1 + 2}{4}) v^{2}$
$g l \sin θ = \frac{3}{4} v^{2}$
$v^{2} = \frac{4 g l \sin θ}{3}$
$v = \sqrt{\frac{4 g l \sin θ}{3}}$
$v = \sqrt{\frac{4 \times 9.8 \times 60 \times 10^{- 2} \times \sin 30^{\circ}}{3}}$
$v = 1.92$
$v ≃ 2 m / s$

TS- EAMCET.11.09.2020

Rotational Motion

150405 A solid sphere of mass $2 kg$ rolls on a smooth horizontal surface at $10 m / s$ . It then rolls up a smooth inclined plane of inclination $30^{\circ}$ with the horizontal. The height attained by the sphere before it stops is [take $g = 10 m / s^{2}$ ]

1

70 cm

2

701 cm

3

7.0 m

4

70 m

Explanation:

C Given,
Mass of solid sphere $= 2 kg$
Velocity $v) = 10 m / s$
Angle $θ) = 30^{\circ}$
original image
If sphere is at point $P$ , sphere has kinetic energy and rotational energy and when it is at point $Q$ , it have only potential energy.
From law of conservation of energy,
$\frac{1}{2} {mv}^{2} + \frac{1}{2} I ω^{2} = mgh$
$\frac{1}{2} {mv}^{2} + \frac{1}{2} \times \frac{2}{5} {mr}^{2} \times \frac{v^{2}}{r^{2}} = mgh [\begin{array}{l} ∴ I = \frac{2}{5} {mr}^{2} \\ and ω = \frac{v}{r} \end{array}]$
${mv}^{2} [\frac{1}{2} + \frac{1}{5}] = mgh$
$\frac{7 v^{2}}{10} = gh$
Put the value of $v$ and $g$ ,
$h = \frac{7 \times (10)^{2}}{10 \times 10}$
$h = 7 m$

TS- EAMCET-11.09.2020

Rotational Motion

150406 A solid sphere and a solid cylinder, each of mass $M$ and radius $R$ are rolling with a linear speed on a flat surface without slipping. Let $L_{1}$ be magnitude of the angular momentum of the sphere with respect to a fixed point along the path of the sphere. Likewise $L_{2}$ be the magnitude of angular momentum of the cylinder with respect to the same fixed point along its path. The ratio $\frac{L_{1}}{L_{2}}$ is

1

\frac{14}{15}

2

\frac{4}{5}

3

\frac{2}{5}

4

\frac{7}{15}

Explanation:

A Given,
Mass of solid sphere and cylinder $= M$
Radius of solid sphere and cylinder $= R$
$L_{1}$ and $L_{2}$ are the magnitude of the angular momentum of the sphere and cylinder.

For linear motion $(L_{1}) = MvR$
$= M (R ω) \times R (∴ v = R ω$
$= {MR}^{2} \times ω$
For rotational motion $(L_{2}) = I ω$
So,
Angular momentum about origin ' $O$ ' $(L_{o}) = L_{1} + L_{2}$
$= {MR}^{2} ω + I ω$
$= ω ({MR}^{2} + I$
We know,
Moment of inertia for solid sphere, $I = \frac{2}{5} {MR}^{2}$
And for solid cylinder, $I = \frac{1}{2} {MR}^{2}$
So, ${(L_{1})}_{about 'O'} = ({MR}^{2} + \frac{2}{5} {MR}^{2}) ω = \frac{7}{5} {MR}^{2} ω$
And ${(L_{2})}_{about ' O^{'}} = ({MR}^{2} + \frac{1}{2} {MR}^{2}) ω = \frac{3}{2} {MR}^{2} ω$
So,
$\frac{L_{1}}{L_{2}} = \frac{\frac{7}{5} {MR}^{2} ω}{\frac{3}{2} {MR}^{2} ω}$
$\frac{L_{1}}{L_{2}} = \frac{14}{15}$

TS- EAMCET-10.09.2020

Rotational Motion

150407 A solid sphere of mass $2 kg$ rolls on a smooth horizontal surface at $10 m / s$ and then rolls up a smooth inclined plane of inclination $30^{\circ}$ with the horizontal. The maximum height reached by the sphere is $(g = 9.8 m / s^{2})$

1

10 m

2

4.9 m

3

14.2 m

4

7.1 m

Explanation:

D Given -
Mass of solid sphere $m) = 2 kg$
Velocity $v) = 10 m / s$

When the sphere is at point $A$ , it possesses kinetic energy and rotational energy and when it is at point $B$ it possesses only potential energy.
So from the law of conservation of energy -
$\frac{1}{2} {mv}^{2} + \frac{1}{2} I ω^{2} = mgh$
$Or \frac{1}{2} {mv}^{2} + \frac{1}{2} {mk}^{2} \times \frac{v^{2}}{R^{2}} = mgh$
$Or \frac{1}{2} {mv}^{2} [1 + \frac{k^{2}}{R^{2}}] = mgh$
$Or$
$\frac{1}{2} \times (10)^{2} [1 + \frac{2}{5}] = 9.8 \times h {$ For solid sphere, $\frac{k^{2}}{R^{2}} = \frac{2}{5}}$
$\frac{1}{2} \times 100 \times \frac{7}{5} = 9.8 h [\begin{array}{l} 70 = 9.8 h \\ h = \frac{700}{98} = 7.1 \end{array}]$
$h = 7.1 m$

Manipal UGET -2020

Rotational Motion

150408 A solid cylinder and a hollow cylinder both of the same mass and same external diameter are released from the same height at the same time on an inclined plane. Both roll down without slipping. Which one will reach the bottom first?

1 Solid cylinder

2 Both together

3 One with higher density

4 Hollow cylinder

Explanation:

A For a rolling body, acceleration is given by
$a = \frac{g \sin θ}{(1 + \frac{k^{2}}{R^{2}})}$
For solid sphere $I = \frac{2}{5} {MR}^{2}$
If radius of gyration is same for both cylinder then
${Mk}^{2} = \frac{2}{5} {MR}^{2}$
$\frac{k^{2}}{R^{2}} = \frac{2}{5}$
$a_{solid} = \frac{5}{7} g \sin θ = 0.71 g \sin θ$
For hollow sphere $I = \frac{1}{2} {MR}^{2}$
${Mk}^{2} = \frac{1}{2} {MR}^{2} \Rightarrow \frac{k^{2}}{R^{2}} = \frac{1}{2}$
$a_{hollow} = \frac{2}{3} g \sin θ = 0.66 g \sin θ$
So,
$a_{solid} > a_{hollow}$
Hence, solid will reach the bottom first.

BITSAT-2020

Rotational Motion

150404 A solid cylinder is released from rest from the top of an inclined plane of inclination $30^{\circ}$ and length $60 cm$ . If the cylinder rolls without slipping, then the speed when it reaches the bottom is

1

1.5 m / s

2

2.0 m / s

3

3.0 m / s

4

6.0 m / s

Explanation:

B Given,
$θ = 30^{\circ}, l = 60 cm = 60 \times 10^{- 2} m$
By work energy theorem,
$mg (l \sin θ) = \frac{1}{2} I ω^{2} + \frac{1}{2} {mv}^{2}$
$mg (l \sin θ) = \frac{1}{2} (\frac{1}{2} {mr}^{2}) {(\frac{v}{r})}^{2} + \frac{1}{2} {mv}^{2} (∵ v = r ω$
$mg (l \sin θ) = (\frac{1}{4} + \frac{1}{2}) {mv}^{2}$
$g l \sin θ = (\frac{1 + 2}{4}) v^{2}$
$g l \sin θ = \frac{3}{4} v^{2}$
$v^{2} = \frac{4 g l \sin θ}{3}$
$v = \sqrt{\frac{4 g l \sin θ}{3}}$
$v = \sqrt{\frac{4 \times 9.8 \times 60 \times 10^{- 2} \times \sin 30^{\circ}}{3}}$
$v = 1.92$
$v ≃ 2 m / s$

TS- EAMCET.11.09.2020

Rotational Motion

150405 A solid sphere of mass $2 kg$ rolls on a smooth horizontal surface at $10 m / s$ . It then rolls up a smooth inclined plane of inclination $30^{\circ}$ with the horizontal. The height attained by the sphere before it stops is [take $g = 10 m / s^{2}$ ]

1

70 cm

2

701 cm

3

7.0 m

4

70 m

Explanation:

C Given,
Mass of solid sphere $= 2 kg$
Velocity $v) = 10 m / s$
Angle $θ) = 30^{\circ}$
original image
If sphere is at point $P$ , sphere has kinetic energy and rotational energy and when it is at point $Q$ , it have only potential energy.
From law of conservation of energy,
$\frac{1}{2} {mv}^{2} + \frac{1}{2} I ω^{2} = mgh$
$\frac{1}{2} {mv}^{2} + \frac{1}{2} \times \frac{2}{5} {mr}^{2} \times \frac{v^{2}}{r^{2}} = mgh [\begin{array}{l} ∴ I = \frac{2}{5} {mr}^{2} \\ and ω = \frac{v}{r} \end{array}]$
${mv}^{2} [\frac{1}{2} + \frac{1}{5}] = mgh$
$\frac{7 v^{2}}{10} = gh$
Put the value of $v$ and $g$ ,
$h = \frac{7 \times (10)^{2}}{10 \times 10}$
$h = 7 m$

TS- EAMCET-11.09.2020

Rotational Motion

150406 A solid sphere and a solid cylinder, each of mass $M$ and radius $R$ are rolling with a linear speed on a flat surface without slipping. Let $L_{1}$ be magnitude of the angular momentum of the sphere with respect to a fixed point along the path of the sphere. Likewise $L_{2}$ be the magnitude of angular momentum of the cylinder with respect to the same fixed point along its path. The ratio $\frac{L_{1}}{L_{2}}$ is

1

\frac{14}{15}

2

\frac{4}{5}

3

\frac{2}{5}

4

\frac{7}{15}

Explanation:

A Given,
Mass of solid sphere and cylinder $= M$
Radius of solid sphere and cylinder $= R$
$L_{1}$ and $L_{2}$ are the magnitude of the angular momentum of the sphere and cylinder.

For linear motion $(L_{1}) = MvR$
$= M (R ω) \times R (∴ v = R ω$
$= {MR}^{2} \times ω$
For rotational motion $(L_{2}) = I ω$
So,
Angular momentum about origin ' $O$ ' $(L_{o}) = L_{1} + L_{2}$
$= {MR}^{2} ω + I ω$
$= ω ({MR}^{2} + I$
We know,
Moment of inertia for solid sphere, $I = \frac{2}{5} {MR}^{2}$
And for solid cylinder, $I = \frac{1}{2} {MR}^{2}$
So, ${(L_{1})}_{about 'O'} = ({MR}^{2} + \frac{2}{5} {MR}^{2}) ω = \frac{7}{5} {MR}^{2} ω$
And ${(L_{2})}_{about ' O^{'}} = ({MR}^{2} + \frac{1}{2} {MR}^{2}) ω = \frac{3}{2} {MR}^{2} ω$
So,
$\frac{L_{1}}{L_{2}} = \frac{\frac{7}{5} {MR}^{2} ω}{\frac{3}{2} {MR}^{2} ω}$
$\frac{L_{1}}{L_{2}} = \frac{14}{15}$

TS- EAMCET-10.09.2020

Rotational Motion

150407 A solid sphere of mass $2 kg$ rolls on a smooth horizontal surface at $10 m / s$ and then rolls up a smooth inclined plane of inclination $30^{\circ}$ with the horizontal. The maximum height reached by the sphere is $(g = 9.8 m / s^{2})$

1

10 m

2

4.9 m

3

14.2 m

4

7.1 m

Explanation:

D Given -
Mass of solid sphere $m) = 2 kg$
Velocity $v) = 10 m / s$

When the sphere is at point $A$ , it possesses kinetic energy and rotational energy and when it is at point $B$ it possesses only potential energy.
So from the law of conservation of energy -
$\frac{1}{2} {mv}^{2} + \frac{1}{2} I ω^{2} = mgh$
$Or \frac{1}{2} {mv}^{2} + \frac{1}{2} {mk}^{2} \times \frac{v^{2}}{R^{2}} = mgh$
$Or \frac{1}{2} {mv}^{2} [1 + \frac{k^{2}}{R^{2}}] = mgh$
$Or$
$\frac{1}{2} \times (10)^{2} [1 + \frac{2}{5}] = 9.8 \times h {$ For solid sphere, $\frac{k^{2}}{R^{2}} = \frac{2}{5}}$
$\frac{1}{2} \times 100 \times \frac{7}{5} = 9.8 h [\begin{array}{l} 70 = 9.8 h \\ h = \frac{700}{98} = 7.1 \end{array}]$
$h = 7.1 m$

Manipal UGET -2020

Rotational Motion

150408 A solid cylinder and a hollow cylinder both of the same mass and same external diameter are released from the same height at the same time on an inclined plane. Both roll down without slipping. Which one will reach the bottom first?

1 Solid cylinder

2 Both together

3 One with higher density

4 Hollow cylinder

Explanation:

A For a rolling body, acceleration is given by
$a = \frac{g \sin θ}{(1 + \frac{k^{2}}{R^{2}})}$
For solid sphere $I = \frac{2}{5} {MR}^{2}$
If radius of gyration is same for both cylinder then
${Mk}^{2} = \frac{2}{5} {MR}^{2}$
$\frac{k^{2}}{R^{2}} = \frac{2}{5}$
$a_{solid} = \frac{5}{7} g \sin θ = 0.71 g \sin θ$
For hollow sphere $I = \frac{1}{2} {MR}^{2}$
${Mk}^{2} = \frac{1}{2} {MR}^{2} \Rightarrow \frac{k^{2}}{R^{2}} = \frac{1}{2}$
$a_{hollow} = \frac{2}{3} g \sin θ = 0.66 g \sin θ$
So,
$a_{solid} > a_{hollow}$
Hence, solid will reach the bottom first.

BITSAT-2020