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PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365977 A force $F$ is applied on a sphere as shown in the figure. For pure accelerated rolling, the force of friction will be towards
supporting img

1 Forward direction

2 Backward direction

3 Normal to the surface

4 Zero

Explanation:

supporting img
Here, $f_{s}$ will be in the forward direction for sphere. $F$ and $f_{s}$ both will contribute in linear and angular accelerations.
$(F + f_{s}) = M \times a (1)$
$(F - f_{s}) R = I α (2)$
For pure acceleration rolling
$a = R α (3)$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365978 A spherical uniform body of radius $R$ , mass $M$ and moment of inertia $I$ rolls down (without slipping) on an inclined plane making an angle $θ$ with the horizontal. then its acceleration is :-

1

\frac{g \sin θ}{1 - I / M R}

2

\frac{g \sin θ}{1 + I / M R^{2}}

3

\frac{g \sin θ}{1 - M R^{2} / I}

4

\frac{g \sin θ}{1 - I / M R}

Explanation:

$a = \frac{g \sin θ}{1 + \frac{I_{C M}}{m R^{2}}}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365979 In a four wheel drive vehicle, the force of friction shall be :

1 Forward on all four wheels

2 Always backward on all four wheels

3 Forward on front wheels and backward on back wheels

4 Backward on front wheels and forward on back wheels

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365980 A disc of radius $\frac{1}{2} m$ has $ω = 2 r a d / s$ and speed $v_{C} = 2 m / s .$ The speed of the plank for pure rolling is
supporting img

1

1 m / s

2

5 m / s

3

7 m / s

4

3 m / s

Explanation:

As disc if rolling the speed of point of contact of disc with plank should be equal to velocity of the plank
$V_{plank} = v_{disc} + ω R$
$= 2 + (2 \times \frac{1}{2}) = 3 m / s$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365977 A force $F$ is applied on a sphere as shown in the figure. For pure accelerated rolling, the force of friction will be towards
supporting img

1 Forward direction

2 Backward direction

3 Normal to the surface

4 Zero

Explanation:

supporting img
Here, $f_{s}$ will be in the forward direction for sphere. $F$ and $f_{s}$ both will contribute in linear and angular accelerations.
$(F + f_{s}) = M \times a (1)$
$(F - f_{s}) R = I α (2)$
For pure acceleration rolling
$a = R α (3)$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365978 A spherical uniform body of radius $R$ , mass $M$ and moment of inertia $I$ rolls down (without slipping) on an inclined plane making an angle $θ$ with the horizontal. then its acceleration is :-

1

\frac{g \sin θ}{1 - I / M R}

2

\frac{g \sin θ}{1 + I / M R^{2}}

3

\frac{g \sin θ}{1 - M R^{2} / I}

4

\frac{g \sin θ}{1 - I / M R}

Explanation:

$a = \frac{g \sin θ}{1 + \frac{I_{C M}}{m R^{2}}}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365979 In a four wheel drive vehicle, the force of friction shall be :

1 Forward on all four wheels

2 Always backward on all four wheels

3 Forward on front wheels and backward on back wheels

4 Backward on front wheels and forward on back wheels

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365980 A disc of radius $\frac{1}{2} m$ has $ω = 2 r a d / s$ and speed $v_{C} = 2 m / s .$ The speed of the plank for pure rolling is
supporting img

1

1 m / s

2

5 m / s

3

7 m / s

4

3 m / s

Explanation:

As disc if rolling the speed of point of contact of disc with plank should be equal to velocity of the plank
$V_{plank} = v_{disc} + ω R$
$= 2 + (2 \times \frac{1}{2}) = 3 m / s$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365977 A force $F$ is applied on a sphere as shown in the figure. For pure accelerated rolling, the force of friction will be towards
supporting img

1 Forward direction

2 Backward direction

3 Normal to the surface

4 Zero

Explanation:

supporting img
Here, $f_{s}$ will be in the forward direction for sphere. $F$ and $f_{s}$ both will contribute in linear and angular accelerations.
$(F + f_{s}) = M \times a (1)$
$(F - f_{s}) R = I α (2)$
For pure acceleration rolling
$a = R α (3)$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365978 A spherical uniform body of radius $R$ , mass $M$ and moment of inertia $I$ rolls down (without slipping) on an inclined plane making an angle $θ$ with the horizontal. then its acceleration is :-

1

\frac{g \sin θ}{1 - I / M R}

2

\frac{g \sin θ}{1 + I / M R^{2}}

3

\frac{g \sin θ}{1 - M R^{2} / I}

4

\frac{g \sin θ}{1 - I / M R}

Explanation:

$a = \frac{g \sin θ}{1 + \frac{I_{C M}}{m R^{2}}}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365979 In a four wheel drive vehicle, the force of friction shall be :

1 Forward on all four wheels

2 Always backward on all four wheels

3 Forward on front wheels and backward on back wheels

4 Backward on front wheels and forward on back wheels

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365980 A disc of radius $\frac{1}{2} m$ has $ω = 2 r a d / s$ and speed $v_{C} = 2 m / s .$ The speed of the plank for pure rolling is
supporting img

1

1 m / s

2

5 m / s

3

7 m / s

4

3 m / s

Explanation:

As disc if rolling the speed of point of contact of disc with plank should be equal to velocity of the plank
$V_{plank} = v_{disc} + ω R$
$= 2 + (2 \times \frac{1}{2}) = 3 m / s$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365977 A force $F$ is applied on a sphere as shown in the figure. For pure accelerated rolling, the force of friction will be towards
supporting img

1 Forward direction

2 Backward direction

3 Normal to the surface

4 Zero

Explanation:

supporting img
Here, $f_{s}$ will be in the forward direction for sphere. $F$ and $f_{s}$ both will contribute in linear and angular accelerations.
$(F + f_{s}) = M \times a (1)$
$(F - f_{s}) R = I α (2)$
For pure acceleration rolling
$a = R α (3)$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365978 A spherical uniform body of radius $R$ , mass $M$ and moment of inertia $I$ rolls down (without slipping) on an inclined plane making an angle $θ$ with the horizontal. then its acceleration is :-

1

\frac{g \sin θ}{1 - I / M R}

2

\frac{g \sin θ}{1 + I / M R^{2}}

3

\frac{g \sin θ}{1 - M R^{2} / I}

4

\frac{g \sin θ}{1 - I / M R}

Explanation:

$a = \frac{g \sin θ}{1 + \frac{I_{C M}}{m R^{2}}}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365979 In a four wheel drive vehicle, the force of friction shall be :

1 Forward on all four wheels

2 Always backward on all four wheels

3 Forward on front wheels and backward on back wheels

4 Backward on front wheels and forward on back wheels

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365980 A disc of radius $\frac{1}{2} m$ has $ω = 2 r a d / s$ and speed $v_{C} = 2 m / s .$ The speed of the plank for pure rolling is
supporting img

1

1 m / s

2

5 m / s

3

7 m / s

4

3 m / s

Explanation:

As disc if rolling the speed of point of contact of disc with plank should be equal to velocity of the plank
$V_{plank} = v_{disc} + ω R$
$= 2 + (2 \times \frac{1}{2}) = 3 m / s$