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PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365989 A uniform disc of mass $M$ and radius $R$ , is resting on a table on its rim. The coeeficient of friction between disc and table is $μ$ . Now the disc is pulled with a force $F$ as shown in the figure. What is the maximum value of $F$ for which the disc rolls without slipping?
supporting img

1

μ M g

2

2 μ M g

3

3 μ M g

4

4 μ M g

Explanation:

supporting img
Let $a$ be the acceleration of the centre of mass of disc. Then
$M a = F - f (1)$
(1) If there is no slipping, angular acceleration of the disc $α = \frac{a}{R} (2)$
(2) Now torque of the disc $τ = I α = R f$
$or τ = (\frac{1}{2} M R^{2}) α = R f [∵ For disc I = \frac{1}{2} M R^{2}]$
$∴ (\frac{1}{2} M R^{2}) (\frac{a}{R}) = R f \Rightarrow M a = 2 f$ (Using $(2)$ )
Substituting this in eq. (1), we get
$2 f = F - f or f = \frac{F}{3}$
Since there is no slipping,
$∴ f \leq μ M g \Rightarrow F \leq 3 μ M g; ∴ F_{max} = 3 μ M g$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365990 A uniform solid sphere of mass $m$ is being pulled on a horizontal surface with force $F$ parallel to the surface and applied at its topmost point. If the acceleration of the sphere is $a$ & it is in pure rolling on the surface, the value of $F$ is

1

m a

2

\frac{3}{2} m a

3

\frac{7}{10} ma

4

\frac{10}{7} m a

Explanation:

supporting img
Writing torque equation about an axis passing through A, perpendicular to the plane of paper.
$\Rightarrow F \times 2 R = (\frac{2}{5} m R^{2} + m R^{2}) \times α (1)$
$a = R α (2)$
From eq’s (1) & (2)
$\Rightarrow F = \frac{7}{10} m a$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365991 A rolling body is rolling without slipping on a moving plank. Relation of $v$ and $v_{1}$ is ...A... Here, $A$ referes to
supporting img

1

v_{1} - V = R ω

2

v_{1} + V = R ω

3

v_{1} + V = - R ω

4 None of these

Explanation:

For pure rolling of sphere relative motion of point of contact of sphere and plank should be zero. For this the point of contact has s velocity equal to the velocity of plank. So, $v_{1} = V + R ω$ .

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365992 A wheel of a bullock cart is rolling on a level road as shown in the figure below. If its linear speed is $v$ in the direction shown, which one of the following options is correct ( $P$ and $Q$ are any highest and lowest points on the wheel, respectively)?
supporting img

1 Point

P

moves slower than point

Q

2 Point

P

moves faster than point

Q

3 Both the points

P

and

Q

move with equal speed

4 Point

P

has zero speed

Explanation:

Here we assume that the wheel is in pure rolling. In the case of pure rolling,
supporting img
The topmost point will have velocity $2 v$ while point $Q$ $i . e .$ lowest point will have zero velocity. Hence point $P$ moves faster than point $Q$ .

NEET - 2024

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365993 A ring rolls down two different inclined planes of same height, but of different inclinations. In both cases

1 Speed will be different, but time of descent will be same

2 Speed will be same, but time of descent will be different

3 Speed and time of descent both are different

4 Speed and time of descent will be same

Explanation:

In pure rolling, mechanical energy remain conserved. Therefore, when heights of inclines are equal, speed of ring will be same in both the cases. But as acceleration $a α \sin θ$ , therefore acceleration and time to descend will be different.

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365989 A uniform disc of mass $M$ and radius $R$ , is resting on a table on its rim. The coeeficient of friction between disc and table is $μ$ . Now the disc is pulled with a force $F$ as shown in the figure. What is the maximum value of $F$ for which the disc rolls without slipping?
supporting img

1

μ M g

2

2 μ M g

3

3 μ M g

4

4 μ M g

Explanation:

supporting img
Let $a$ be the acceleration of the centre of mass of disc. Then
$M a = F - f (1)$
(1) If there is no slipping, angular acceleration of the disc $α = \frac{a}{R} (2)$
(2) Now torque of the disc $τ = I α = R f$
$or τ = (\frac{1}{2} M R^{2}) α = R f [∵ For disc I = \frac{1}{2} M R^{2}]$
$∴ (\frac{1}{2} M R^{2}) (\frac{a}{R}) = R f \Rightarrow M a = 2 f$ (Using $(2)$ )
Substituting this in eq. (1), we get
$2 f = F - f or f = \frac{F}{3}$
Since there is no slipping,
$∴ f \leq μ M g \Rightarrow F \leq 3 μ M g; ∴ F_{max} = 3 μ M g$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365990 A uniform solid sphere of mass $m$ is being pulled on a horizontal surface with force $F$ parallel to the surface and applied at its topmost point. If the acceleration of the sphere is $a$ & it is in pure rolling on the surface, the value of $F$ is

1

m a

2

\frac{3}{2} m a

3

\frac{7}{10} ma

4

\frac{10}{7} m a

Explanation:

supporting img
Writing torque equation about an axis passing through A, perpendicular to the plane of paper.
$\Rightarrow F \times 2 R = (\frac{2}{5} m R^{2} + m R^{2}) \times α (1)$
$a = R α (2)$
From eq’s (1) & (2)
$\Rightarrow F = \frac{7}{10} m a$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365991 A rolling body is rolling without slipping on a moving plank. Relation of $v$ and $v_{1}$ is ...A... Here, $A$ referes to
supporting img

1

v_{1} - V = R ω

2

v_{1} + V = R ω

3

v_{1} + V = - R ω

4 None of these

Explanation:

For pure rolling of sphere relative motion of point of contact of sphere and plank should be zero. For this the point of contact has s velocity equal to the velocity of plank. So, $v_{1} = V + R ω$ .

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365992 A wheel of a bullock cart is rolling on a level road as shown in the figure below. If its linear speed is $v$ in the direction shown, which one of the following options is correct ( $P$ and $Q$ are any highest and lowest points on the wheel, respectively)?
supporting img

1 Point

P

moves slower than point

Q

2 Point

P

moves faster than point

Q

3 Both the points

P

and

Q

move with equal speed

4 Point

P

has zero speed

Explanation:

Here we assume that the wheel is in pure rolling. In the case of pure rolling,
supporting img
The topmost point will have velocity $2 v$ while point $Q$ $i . e .$ lowest point will have zero velocity. Hence point $P$ moves faster than point $Q$ .

NEET - 2024

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365993 A ring rolls down two different inclined planes of same height, but of different inclinations. In both cases

1 Speed will be different, but time of descent will be same

2 Speed will be same, but time of descent will be different

3 Speed and time of descent both are different

4 Speed and time of descent will be same

Explanation:

In pure rolling, mechanical energy remain conserved. Therefore, when heights of inclines are equal, speed of ring will be same in both the cases. But as acceleration $a α \sin θ$ , therefore acceleration and time to descend will be different.

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365989 A uniform disc of mass $M$ and radius $R$ , is resting on a table on its rim. The coeeficient of friction between disc and table is $μ$ . Now the disc is pulled with a force $F$ as shown in the figure. What is the maximum value of $F$ for which the disc rolls without slipping?
supporting img

1

μ M g

2

2 μ M g

3

3 μ M g

4

4 μ M g

Explanation:

supporting img
Let $a$ be the acceleration of the centre of mass of disc. Then
$M a = F - f (1)$
(1) If there is no slipping, angular acceleration of the disc $α = \frac{a}{R} (2)$
(2) Now torque of the disc $τ = I α = R f$
$or τ = (\frac{1}{2} M R^{2}) α = R f [∵ For disc I = \frac{1}{2} M R^{2}]$
$∴ (\frac{1}{2} M R^{2}) (\frac{a}{R}) = R f \Rightarrow M a = 2 f$ (Using $(2)$ )
Substituting this in eq. (1), we get
$2 f = F - f or f = \frac{F}{3}$
Since there is no slipping,
$∴ f \leq μ M g \Rightarrow F \leq 3 μ M g; ∴ F_{max} = 3 μ M g$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365990 A uniform solid sphere of mass $m$ is being pulled on a horizontal surface with force $F$ parallel to the surface and applied at its topmost point. If the acceleration of the sphere is $a$ & it is in pure rolling on the surface, the value of $F$ is

1

m a

2

\frac{3}{2} m a

3

\frac{7}{10} ma

4

\frac{10}{7} m a

Explanation:

supporting img
Writing torque equation about an axis passing through A, perpendicular to the plane of paper.
$\Rightarrow F \times 2 R = (\frac{2}{5} m R^{2} + m R^{2}) \times α (1)$
$a = R α (2)$
From eq’s (1) & (2)
$\Rightarrow F = \frac{7}{10} m a$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365991 A rolling body is rolling without slipping on a moving plank. Relation of $v$ and $v_{1}$ is ...A... Here, $A$ referes to
supporting img

1

v_{1} - V = R ω

2

v_{1} + V = R ω

3

v_{1} + V = - R ω

4 None of these

Explanation:

For pure rolling of sphere relative motion of point of contact of sphere and plank should be zero. For this the point of contact has s velocity equal to the velocity of plank. So, $v_{1} = V + R ω$ .

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365992 A wheel of a bullock cart is rolling on a level road as shown in the figure below. If its linear speed is $v$ in the direction shown, which one of the following options is correct ( $P$ and $Q$ are any highest and lowest points on the wheel, respectively)?
supporting img

1 Point

P

moves slower than point

Q

2 Point

P

moves faster than point

Q

3 Both the points

P

and

Q

move with equal speed

4 Point

P

has zero speed

Explanation:

Here we assume that the wheel is in pure rolling. In the case of pure rolling,
supporting img
The topmost point will have velocity $2 v$ while point $Q$ $i . e .$ lowest point will have zero velocity. Hence point $P$ moves faster than point $Q$ .

NEET - 2024

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365993 A ring rolls down two different inclined planes of same height, but of different inclinations. In both cases

1 Speed will be different, but time of descent will be same

2 Speed will be same, but time of descent will be different

3 Speed and time of descent both are different

4 Speed and time of descent will be same

Explanation:

In pure rolling, mechanical energy remain conserved. Therefore, when heights of inclines are equal, speed of ring will be same in both the cases. But as acceleration $a α \sin θ$ , therefore acceleration and time to descend will be different.

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365989 A uniform disc of mass $M$ and radius $R$ , is resting on a table on its rim. The coeeficient of friction between disc and table is $μ$ . Now the disc is pulled with a force $F$ as shown in the figure. What is the maximum value of $F$ for which the disc rolls without slipping?
supporting img

1

μ M g

2

2 μ M g

3

3 μ M g

4

4 μ M g

Explanation:

supporting img
Let $a$ be the acceleration of the centre of mass of disc. Then
$M a = F - f (1)$
(1) If there is no slipping, angular acceleration of the disc $α = \frac{a}{R} (2)$
(2) Now torque of the disc $τ = I α = R f$
$or τ = (\frac{1}{2} M R^{2}) α = R f [∵ For disc I = \frac{1}{2} M R^{2}]$
$∴ (\frac{1}{2} M R^{2}) (\frac{a}{R}) = R f \Rightarrow M a = 2 f$ (Using $(2)$ )
Substituting this in eq. (1), we get
$2 f = F - f or f = \frac{F}{3}$
Since there is no slipping,
$∴ f \leq μ M g \Rightarrow F \leq 3 μ M g; ∴ F_{max} = 3 μ M g$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365990 A uniform solid sphere of mass $m$ is being pulled on a horizontal surface with force $F$ parallel to the surface and applied at its topmost point. If the acceleration of the sphere is $a$ & it is in pure rolling on the surface, the value of $F$ is

1

m a

2

\frac{3}{2} m a

3

\frac{7}{10} ma

4

\frac{10}{7} m a

Explanation:

supporting img
Writing torque equation about an axis passing through A, perpendicular to the plane of paper.
$\Rightarrow F \times 2 R = (\frac{2}{5} m R^{2} + m R^{2}) \times α (1)$
$a = R α (2)$
From eq’s (1) & (2)
$\Rightarrow F = \frac{7}{10} m a$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365991 A rolling body is rolling without slipping on a moving plank. Relation of $v$ and $v_{1}$ is ...A... Here, $A$ referes to
supporting img

1

v_{1} - V = R ω

2

v_{1} + V = R ω

3

v_{1} + V = - R ω

4 None of these

Explanation:

For pure rolling of sphere relative motion of point of contact of sphere and plank should be zero. For this the point of contact has s velocity equal to the velocity of plank. So, $v_{1} = V + R ω$ .

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365992 A wheel of a bullock cart is rolling on a level road as shown in the figure below. If its linear speed is $v$ in the direction shown, which one of the following options is correct ( $P$ and $Q$ are any highest and lowest points on the wheel, respectively)?
supporting img

1 Point

P

moves slower than point

Q

2 Point

P

moves faster than point

Q

3 Both the points

P

and

Q

move with equal speed

4 Point

P

has zero speed

Explanation:

Here we assume that the wheel is in pure rolling. In the case of pure rolling,
supporting img
The topmost point will have velocity $2 v$ while point $Q$ $i . e .$ lowest point will have zero velocity. Hence point $P$ moves faster than point $Q$ .

NEET - 2024

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365993 A ring rolls down two different inclined planes of same height, but of different inclinations. In both cases

1 Speed will be different, but time of descent will be same

2 Speed will be same, but time of descent will be different

3 Speed and time of descent both are different

4 Speed and time of descent will be same

Explanation:

In pure rolling, mechanical energy remain conserved. Therefore, when heights of inclines are equal, speed of ring will be same in both the cases. But as acceleration $a α \sin θ$ , therefore acceleration and time to descend will be different.

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365989 A uniform disc of mass $M$ and radius $R$ , is resting on a table on its rim. The coeeficient of friction between disc and table is $μ$ . Now the disc is pulled with a force $F$ as shown in the figure. What is the maximum value of $F$ for which the disc rolls without slipping?
supporting img

1

μ M g

2

2 μ M g

3

3 μ M g

4

4 μ M g

Explanation:

supporting img
Let $a$ be the acceleration of the centre of mass of disc. Then
$M a = F - f (1)$
(1) If there is no slipping, angular acceleration of the disc $α = \frac{a}{R} (2)$
(2) Now torque of the disc $τ = I α = R f$
$or τ = (\frac{1}{2} M R^{2}) α = R f [∵ For disc I = \frac{1}{2} M R^{2}]$
$∴ (\frac{1}{2} M R^{2}) (\frac{a}{R}) = R f \Rightarrow M a = 2 f$ (Using $(2)$ )
Substituting this in eq. (1), we get
$2 f = F - f or f = \frac{F}{3}$
Since there is no slipping,
$∴ f \leq μ M g \Rightarrow F \leq 3 μ M g; ∴ F_{max} = 3 μ M g$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365990 A uniform solid sphere of mass $m$ is being pulled on a horizontal surface with force $F$ parallel to the surface and applied at its topmost point. If the acceleration of the sphere is $a$ & it is in pure rolling on the surface, the value of $F$ is

1

m a

2

\frac{3}{2} m a

3

\frac{7}{10} ma

4

\frac{10}{7} m a

Explanation:

supporting img
Writing torque equation about an axis passing through A, perpendicular to the plane of paper.
$\Rightarrow F \times 2 R = (\frac{2}{5} m R^{2} + m R^{2}) \times α (1)$
$a = R α (2)$
From eq’s (1) & (2)
$\Rightarrow F = \frac{7}{10} m a$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365991 A rolling body is rolling without slipping on a moving plank. Relation of $v$ and $v_{1}$ is ...A... Here, $A$ referes to
supporting img

1

v_{1} - V = R ω

2

v_{1} + V = R ω

3

v_{1} + V = - R ω

4 None of these

Explanation:

For pure rolling of sphere relative motion of point of contact of sphere and plank should be zero. For this the point of contact has s velocity equal to the velocity of plank. So, $v_{1} = V + R ω$ .

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365992 A wheel of a bullock cart is rolling on a level road as shown in the figure below. If its linear speed is $v$ in the direction shown, which one of the following options is correct ( $P$ and $Q$ are any highest and lowest points on the wheel, respectively)?
supporting img

1 Point

P

moves slower than point

Q

2 Point

P

moves faster than point

Q

3 Both the points

P

and

Q

move with equal speed

4 Point

P

has zero speed

Explanation:

Here we assume that the wheel is in pure rolling. In the case of pure rolling,
supporting img
The topmost point will have velocity $2 v$ while point $Q$ $i . e .$ lowest point will have zero velocity. Hence point $P$ moves faster than point $Q$ .

NEET - 2024

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365993 A ring rolls down two different inclined planes of same height, but of different inclinations. In both cases

1 Speed will be different, but time of descent will be same

2 Speed will be same, but time of descent will be different

3 Speed and time of descent both are different

4 Speed and time of descent will be same

Explanation:

In pure rolling, mechanical energy remain conserved. Therefore, when heights of inclines are equal, speed of ring will be same in both the cases. But as acceleration $a α \sin θ$ , therefore acceleration and time to descend will be different.