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Rotational Motion

150465 A thin uniform circular ring is rolling down an inclined plane of inclination $30^{\circ}$ without slipping. Its linear acceleration along the inclined plane will be

1

\frac{g}{2}

2

\frac{g}{3}

3

\frac{g}{4}

4

\frac{2 g}{3}

Explanation:

C original image
We know that, acceleration on inclined plane in rolling motion-
$a = \frac{g \sin θ}{1 + \frac{K^{2}}{R^{2}}}$
For ring, $\frac{K^{2}}{R^{2}} = 1$
$a = \frac{g \sin 30^{\circ}}{1 + 1}$
$a = \frac{g \times \frac{1}{2}}{2}$
$[∵ \sin 30^{\circ} = \frac{1}{2}]$
$a = \frac{g}{4}$

AIPMT- 1994

Rotational Motion

150466 A very small particle rests on the top of a hemisphere of radius $20 cm$ . The smallest horizontal velocity to be given to it, if it is to leave the hemisphere without sliding down its surface, taking $g = 9.8 {m s}^{- 2}$ is

1

1.4 {ms}^{- 1}

2

2.4 {ms}^{- 1}

3

0.4 {ms}^{- 1}

4

0.7 {ms}^{- 1}

Explanation:

A original image
Given,
$R = 20 cm = 0.2 m$
$g = 9.8 m / s^{2}$
According to question,
When particle is rest on the top of hemisphere-
$mg - N = \frac{mv v^{2}}{R}$
If particle leave the hemisphere without sliding then,
$N = 0$
$∴ mg = \frac{{mv}^{2}}{R}$
$v = \sqrt{Rg}$
$v = \sqrt{0.2 \times 9.8} m / s$
$v = \sqrt{1.96}$
$v = 1.4 m / s$

EAMCET-1993

Rotational Motion

150467 A rod of length $l$ is held vertically stationary with its lower end located at a position $P$ on the horizontal plane. When the rod is released to topple about $P$ , the velocity of the upper end of the rod with which it hits the ground is

1

\sqrt{\frac{g}{l}}

2

\sqrt{3 g l}

3

3 \sqrt{\frac{g}{l}}

4

\sqrt{\frac{3 g}{l}}

Explanation:

B
The rod topples about point $P$ .
So, $MOI$ of $rod (I) = \frac{M l^{2}}{3}$
By law of conservation of energy-
Loss in P.E = gain in K.E
$mg \frac{l}{2} = \frac{1}{2} I ω^{2}$
$mg \frac{l}{2} = \frac{1}{2} \frac{m l^{2}}{3} \times ω^{2} [∵ I = \frac{m l^{2}}{3}]$
$mg \frac{l}{2} = \frac{1}{2} \frac{m l^{2}}{3} \times ω^{2} [∵ I = \frac{m l^{2}}{3}]$
For rotation about point $P$ -
$∴ v = ω l$
$So, v = \sqrt{\frac{3 g}{l}} \times l$
$v = \sqrt{3 g l}$

AP EAMCET(Medical)-2009

Rotational Motion

150443 Two uniform solid spheres having unequal masses and unequal radii are released from rest from the same height on a rough incline. If the spheres roll without slipping,

1 the heavier sphere reaches the bottom first

2 the bigger sphere reaches the bottom first

3 the two spheres reach the bottom together

4 the information given is not sufficient to tell which sphere will reach the bottom first

Explanation:

C For solid sphere-
$a = \frac{g \sin θ}{(1 + \frac{K^{2}}{R^{2}})}$
$\frac{K^{2}}{R^{2}} = \frac{2}{5}$ (Where $K$ is radius of gyration)
So acceleration is not affected in this case because it independent of mass. So, both sphere reach the bottom together.

BITSAT-2007

Rotational Motion

150465 A thin uniform circular ring is rolling down an inclined plane of inclination $30^{\circ}$ without slipping. Its linear acceleration along the inclined plane will be

1

\frac{g}{2}

2

\frac{g}{3}

3

\frac{g}{4}

4

\frac{2 g}{3}

Explanation:

C original image
We know that, acceleration on inclined plane in rolling motion-
$a = \frac{g \sin θ}{1 + \frac{K^{2}}{R^{2}}}$
For ring, $\frac{K^{2}}{R^{2}} = 1$
$a = \frac{g \sin 30^{\circ}}{1 + 1}$
$a = \frac{g \times \frac{1}{2}}{2}$
$[∵ \sin 30^{\circ} = \frac{1}{2}]$
$a = \frac{g}{4}$

AIPMT- 1994

Rotational Motion

150466 A very small particle rests on the top of a hemisphere of radius $20 cm$ . The smallest horizontal velocity to be given to it, if it is to leave the hemisphere without sliding down its surface, taking $g = 9.8 {m s}^{- 2}$ is

1

1.4 {ms}^{- 1}

2

2.4 {ms}^{- 1}

3

0.4 {ms}^{- 1}

4

0.7 {ms}^{- 1}

Explanation:

A original image
Given,
$R = 20 cm = 0.2 m$
$g = 9.8 m / s^{2}$
According to question,
When particle is rest on the top of hemisphere-
$mg - N = \frac{mv v^{2}}{R}$
If particle leave the hemisphere without sliding then,
$N = 0$
$∴ mg = \frac{{mv}^{2}}{R}$
$v = \sqrt{Rg}$
$v = \sqrt{0.2 \times 9.8} m / s$
$v = \sqrt{1.96}$
$v = 1.4 m / s$

EAMCET-1993

Rotational Motion

150467 A rod of length $l$ is held vertically stationary with its lower end located at a position $P$ on the horizontal plane. When the rod is released to topple about $P$ , the velocity of the upper end of the rod with which it hits the ground is

1

\sqrt{\frac{g}{l}}

2

\sqrt{3 g l}

3

3 \sqrt{\frac{g}{l}}

4

\sqrt{\frac{3 g}{l}}

Explanation:

B
The rod topples about point $P$ .
So, $MOI$ of $rod (I) = \frac{M l^{2}}{3}$
By law of conservation of energy-
Loss in P.E = gain in K.E
$mg \frac{l}{2} = \frac{1}{2} I ω^{2}$
$mg \frac{l}{2} = \frac{1}{2} \frac{m l^{2}}{3} \times ω^{2} [∵ I = \frac{m l^{2}}{3}]$
$mg \frac{l}{2} = \frac{1}{2} \frac{m l^{2}}{3} \times ω^{2} [∵ I = \frac{m l^{2}}{3}]$
For rotation about point $P$ -
$∴ v = ω l$
$So, v = \sqrt{\frac{3 g}{l}} \times l$
$v = \sqrt{3 g l}$

AP EAMCET(Medical)-2009

Rotational Motion

150443 Two uniform solid spheres having unequal masses and unequal radii are released from rest from the same height on a rough incline. If the spheres roll without slipping,

1 the heavier sphere reaches the bottom first

2 the bigger sphere reaches the bottom first

3 the two spheres reach the bottom together

4 the information given is not sufficient to tell which sphere will reach the bottom first

Explanation:

C For solid sphere-
$a = \frac{g \sin θ}{(1 + \frac{K^{2}}{R^{2}})}$
$\frac{K^{2}}{R^{2}} = \frac{2}{5}$ (Where $K$ is radius of gyration)
So acceleration is not affected in this case because it independent of mass. So, both sphere reach the bottom together.

BITSAT-2007

Rotational Motion

150465 A thin uniform circular ring is rolling down an inclined plane of inclination $30^{\circ}$ without slipping. Its linear acceleration along the inclined plane will be

1

\frac{g}{2}

2

\frac{g}{3}

3

\frac{g}{4}

4

\frac{2 g}{3}

Explanation:

C original image
We know that, acceleration on inclined plane in rolling motion-
$a = \frac{g \sin θ}{1 + \frac{K^{2}}{R^{2}}}$
For ring, $\frac{K^{2}}{R^{2}} = 1$
$a = \frac{g \sin 30^{\circ}}{1 + 1}$
$a = \frac{g \times \frac{1}{2}}{2}$
$[∵ \sin 30^{\circ} = \frac{1}{2}]$
$a = \frac{g}{4}$

AIPMT- 1994

Rotational Motion

150466 A very small particle rests on the top of a hemisphere of radius $20 cm$ . The smallest horizontal velocity to be given to it, if it is to leave the hemisphere without sliding down its surface, taking $g = 9.8 {m s}^{- 2}$ is

1

1.4 {ms}^{- 1}

2

2.4 {ms}^{- 1}

3

0.4 {ms}^{- 1}

4

0.7 {ms}^{- 1}

Explanation:

A original image
Given,
$R = 20 cm = 0.2 m$
$g = 9.8 m / s^{2}$
According to question,
When particle is rest on the top of hemisphere-
$mg - N = \frac{mv v^{2}}{R}$
If particle leave the hemisphere without sliding then,
$N = 0$
$∴ mg = \frac{{mv}^{2}}{R}$
$v = \sqrt{Rg}$
$v = \sqrt{0.2 \times 9.8} m / s$
$v = \sqrt{1.96}$
$v = 1.4 m / s$

EAMCET-1993

Rotational Motion

150467 A rod of length $l$ is held vertically stationary with its lower end located at a position $P$ on the horizontal plane. When the rod is released to topple about $P$ , the velocity of the upper end of the rod with which it hits the ground is

1

\sqrt{\frac{g}{l}}

2

\sqrt{3 g l}

3

3 \sqrt{\frac{g}{l}}

4

\sqrt{\frac{3 g}{l}}

Explanation:

B
The rod topples about point $P$ .
So, $MOI$ of $rod (I) = \frac{M l^{2}}{3}$
By law of conservation of energy-
Loss in P.E = gain in K.E
$mg \frac{l}{2} = \frac{1}{2} I ω^{2}$
$mg \frac{l}{2} = \frac{1}{2} \frac{m l^{2}}{3} \times ω^{2} [∵ I = \frac{m l^{2}}{3}]$
$mg \frac{l}{2} = \frac{1}{2} \frac{m l^{2}}{3} \times ω^{2} [∵ I = \frac{m l^{2}}{3}]$
For rotation about point $P$ -
$∴ v = ω l$
$So, v = \sqrt{\frac{3 g}{l}} \times l$
$v = \sqrt{3 g l}$

AP EAMCET(Medical)-2009

Rotational Motion

150443 Two uniform solid spheres having unequal masses and unequal radii are released from rest from the same height on a rough incline. If the spheres roll without slipping,

1 the heavier sphere reaches the bottom first

2 the bigger sphere reaches the bottom first

3 the two spheres reach the bottom together

4 the information given is not sufficient to tell which sphere will reach the bottom first

Explanation:

C For solid sphere-
$a = \frac{g \sin θ}{(1 + \frac{K^{2}}{R^{2}})}$
$\frac{K^{2}}{R^{2}} = \frac{2}{5}$ (Where $K$ is radius of gyration)
So acceleration is not affected in this case because it independent of mass. So, both sphere reach the bottom together.

BITSAT-2007

Rotational Motion

150465 A thin uniform circular ring is rolling down an inclined plane of inclination $30^{\circ}$ without slipping. Its linear acceleration along the inclined plane will be

1

\frac{g}{2}

2

\frac{g}{3}

3

\frac{g}{4}

4

\frac{2 g}{3}

Explanation:

C original image
We know that, acceleration on inclined plane in rolling motion-
$a = \frac{g \sin θ}{1 + \frac{K^{2}}{R^{2}}}$
For ring, $\frac{K^{2}}{R^{2}} = 1$
$a = \frac{g \sin 30^{\circ}}{1 + 1}$
$a = \frac{g \times \frac{1}{2}}{2}$
$[∵ \sin 30^{\circ} = \frac{1}{2}]$
$a = \frac{g}{4}$

AIPMT- 1994

Rotational Motion

150466 A very small particle rests on the top of a hemisphere of radius $20 cm$ . The smallest horizontal velocity to be given to it, if it is to leave the hemisphere without sliding down its surface, taking $g = 9.8 {m s}^{- 2}$ is

1

1.4 {ms}^{- 1}

2

2.4 {ms}^{- 1}

3

0.4 {ms}^{- 1}

4

0.7 {ms}^{- 1}

Explanation:

A original image
Given,
$R = 20 cm = 0.2 m$
$g = 9.8 m / s^{2}$
According to question,
When particle is rest on the top of hemisphere-
$mg - N = \frac{mv v^{2}}{R}$
If particle leave the hemisphere without sliding then,
$N = 0$
$∴ mg = \frac{{mv}^{2}}{R}$
$v = \sqrt{Rg}$
$v = \sqrt{0.2 \times 9.8} m / s$
$v = \sqrt{1.96}$
$v = 1.4 m / s$

EAMCET-1993

Rotational Motion

150467 A rod of length $l$ is held vertically stationary with its lower end located at a position $P$ on the horizontal plane. When the rod is released to topple about $P$ , the velocity of the upper end of the rod with which it hits the ground is

1

\sqrt{\frac{g}{l}}

2

\sqrt{3 g l}

3

3 \sqrt{\frac{g}{l}}

4

\sqrt{\frac{3 g}{l}}

Explanation:

B
The rod topples about point $P$ .
So, $MOI$ of $rod (I) = \frac{M l^{2}}{3}$
By law of conservation of energy-
Loss in P.E = gain in K.E
$mg \frac{l}{2} = \frac{1}{2} I ω^{2}$
$mg \frac{l}{2} = \frac{1}{2} \frac{m l^{2}}{3} \times ω^{2} [∵ I = \frac{m l^{2}}{3}]$
$mg \frac{l}{2} = \frac{1}{2} \frac{m l^{2}}{3} \times ω^{2} [∵ I = \frac{m l^{2}}{3}]$
For rotation about point $P$ -
$∴ v = ω l$
$So, v = \sqrt{\frac{3 g}{l}} \times l$
$v = \sqrt{3 g l}$

AP EAMCET(Medical)-2009

Rotational Motion

150443 Two uniform solid spheres having unequal masses and unequal radii are released from rest from the same height on a rough incline. If the spheres roll without slipping,

1 the heavier sphere reaches the bottom first

2 the bigger sphere reaches the bottom first

3 the two spheres reach the bottom together

4 the information given is not sufficient to tell which sphere will reach the bottom first

Explanation:

C For solid sphere-
$a = \frac{g \sin θ}{(1 + \frac{K^{2}}{R^{2}})}$
$\frac{K^{2}}{R^{2}} = \frac{2}{5}$ (Where $K$ is radius of gyration)
So acceleration is not affected in this case because it independent of mass. So, both sphere reach the bottom together.

BITSAT-2007

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