QBManager

Rotational Motion

149922 A constant torque of $1000 N - m$ turns a wheel of moment of inertia $200 kg - m^{2}$ about an axis through its centre. Its angular velocity after 3 second is, in $rad / s$ :

1 1

2 5

3 15

4 10

Explanation:

C Given,
Torque $(τ) = 1000 N - m$
Moment of Inertia (I) $= 200 kg m^{2}$
Time $(t) = 3 \sec$
We know that,
$τ = I α$
$1000 = 200 \times α$
$α = 5 rad / \sec^{2}$
From equation of motion,
$ω = ω_{0} + α t$
$ω = 0 + 5 \times 3$
$ω = 15 rad / \sec$

CPMT-1999

Rotational Motion

149923 The angular momentum of a particle with respect to the origin will not be zero, if

1 the directional line of linear momentum passes through the origin

2 the particle is at the origin (c) the angle between the position vector and linear momentum is

180^{\circ}

3 (d.) the linear momentum vanishes

4 the angle between the position vector and linear momentum is

90^{\circ}

Explanation:

E We know that,
angular momentum, $L = \vec{r} \times \vec{p} = rp \sin θ$
The conditions in which the value of angular momentum is zero, are given below.
(1) When the directional line of liner momentum passes through the origin it means the value of $θ$ will be zero so that $\sin θ = 0$ .
Hence angular momentum (L) will be zero
(2) When the value of position vector is zero, it means the particle is at the origin.
(3) When the value of $θ$ will be $180^{\circ}$ , it means the angle between the position vector and linear momentum is $180^{\circ}$ .
(4) When the value of linear momentum will be zero, it means the linear momentum vanishes.

Kerala CEE 04.07.2022

Rotational Motion

149924 A particle is moving in an elliptical orbit as shown in figure. If $\vec{p}, \vec{L}$ and $\vec{r}$ denote the linear momentum, angular momentum and position vector of the particle (from focus $O$ respectively at a point at $A$ , then the direction of $\vec{α} = \vec{p} \times \vec{L}$ is along.

1 + ve

x

axis

2 -ve

x

axis

3 + ve y axis

4 - ve y axis

Explanation:

A
From figure,
Linear momentum is moving anti-clockwise direction and perpendicular to radius from point A towards +ve $y$ -axis ( p $\hat{j}$ .
From right hand thumb Rule,
Angular momentum $\vec{L}$ is towards +ve z-axis $L \hat{k}$
Now,
$\vec{α} = \vec{p} \times \vec{L}$
$\vec{α} = (p \hat{j}) \times (L \hat{k}$
$= p L (\hat{j} \times \hat{k}$
$= p L \hat{i}$
From properties of cross product of 2 vector, therefore the direction of $α$ is +ve $x$ -axis.

WB JEE 2022

Rotational Motion

149925 If the radius of a spherical object, rotating about its diameter with a time period of 2 second, is reduced to half its actual value, keeping its mass unchanged, its time period becomes (assuming zero external torque)

1 Remains the same

2

6 s

3

0.5 s

4

1 s

Explanation:

C Initial angular momentum of spherical object
$L_{1} = I ω = \frac{2}{5} {MR}^{2} ω$
Final angular momentum of Spherical object
$L_{2} = I ω = \frac{2}{5} M {(\frac{R}{2})}^{2} ω^{'}$
$∵$ External torque is zero (Given)
So, Momentum will be conserved-
$L_{1} = L_{2}$
$\frac{2}{5} {MR}^{2} ω = \frac{2}{5} M {(\frac{R}{2})}^{2} ω^{'}$
$ω R^{2} = \frac{R^{2}}{4} ω^{'}$
$\Rightarrow \frac{1}{T} = \frac{1}{4 T^{'}}$
$\frac{1}{2} = \frac{1}{4 T^{'}} (T = 2 \sec .$
$T^{'} = \frac{1}{2} \sec .$
$T^{'} = 0.5 \sec$

AP EAMCET-03.09.2021

Rotational Motion

149922 A constant torque of $1000 N - m$ turns a wheel of moment of inertia $200 kg - m^{2}$ about an axis through its centre. Its angular velocity after 3 second is, in $rad / s$ :

1 1

2 5

3 15

4 10

Explanation:

C Given,
Torque $(τ) = 1000 N - m$
Moment of Inertia (I) $= 200 kg m^{2}$
Time $(t) = 3 \sec$
We know that,
$τ = I α$
$1000 = 200 \times α$
$α = 5 rad / \sec^{2}$
From equation of motion,
$ω = ω_{0} + α t$
$ω = 0 + 5 \times 3$
$ω = 15 rad / \sec$

CPMT-1999

Rotational Motion

149923 The angular momentum of a particle with respect to the origin will not be zero, if

1 the directional line of linear momentum passes through the origin

2 the particle is at the origin (c) the angle between the position vector and linear momentum is

180^{\circ}

3 (d.) the linear momentum vanishes

4 the angle between the position vector and linear momentum is

90^{\circ}

Explanation:

E We know that,
angular momentum, $L = \vec{r} \times \vec{p} = rp \sin θ$
The conditions in which the value of angular momentum is zero, are given below.
(1) When the directional line of liner momentum passes through the origin it means the value of $θ$ will be zero so that $\sin θ = 0$ .
Hence angular momentum (L) will be zero
(2) When the value of position vector is zero, it means the particle is at the origin.
(3) When the value of $θ$ will be $180^{\circ}$ , it means the angle between the position vector and linear momentum is $180^{\circ}$ .
(4) When the value of linear momentum will be zero, it means the linear momentum vanishes.

Kerala CEE 04.07.2022

Rotational Motion

149924 A particle is moving in an elliptical orbit as shown in figure. If $\vec{p}, \vec{L}$ and $\vec{r}$ denote the linear momentum, angular momentum and position vector of the particle (from focus $O$ respectively at a point at $A$ , then the direction of $\vec{α} = \vec{p} \times \vec{L}$ is along.

1 + ve

x

axis

2 -ve

x

axis

3 + ve y axis

4 - ve y axis

Explanation:

A
From figure,
Linear momentum is moving anti-clockwise direction and perpendicular to radius from point A towards +ve $y$ -axis ( p $\hat{j}$ .
From right hand thumb Rule,
Angular momentum $\vec{L}$ is towards +ve z-axis $L \hat{k}$
Now,
$\vec{α} = \vec{p} \times \vec{L}$
$\vec{α} = (p \hat{j}) \times (L \hat{k}$
$= p L (\hat{j} \times \hat{k}$
$= p L \hat{i}$
From properties of cross product of 2 vector, therefore the direction of $α$ is +ve $x$ -axis.

WB JEE 2022

Rotational Motion

149925 If the radius of a spherical object, rotating about its diameter with a time period of 2 second, is reduced to half its actual value, keeping its mass unchanged, its time period becomes (assuming zero external torque)

1 Remains the same

2

6 s

3

0.5 s

4

1 s

Explanation:

C Initial angular momentum of spherical object
$L_{1} = I ω = \frac{2}{5} {MR}^{2} ω$
Final angular momentum of Spherical object
$L_{2} = I ω = \frac{2}{5} M {(\frac{R}{2})}^{2} ω^{'}$
$∵$ External torque is zero (Given)
So, Momentum will be conserved-
$L_{1} = L_{2}$
$\frac{2}{5} {MR}^{2} ω = \frac{2}{5} M {(\frac{R}{2})}^{2} ω^{'}$
$ω R^{2} = \frac{R^{2}}{4} ω^{'}$
$\Rightarrow \frac{1}{T} = \frac{1}{4 T^{'}}$
$\frac{1}{2} = \frac{1}{4 T^{'}} (T = 2 \sec .$
$T^{'} = \frac{1}{2} \sec .$
$T^{'} = 0.5 \sec$

AP EAMCET-03.09.2021

Rotational Motion

149922 A constant torque of $1000 N - m$ turns a wheel of moment of inertia $200 kg - m^{2}$ about an axis through its centre. Its angular velocity after 3 second is, in $rad / s$ :

1 1

2 5

3 15

4 10

Explanation:

C Given,
Torque $(τ) = 1000 N - m$
Moment of Inertia (I) $= 200 kg m^{2}$
Time $(t) = 3 \sec$
We know that,
$τ = I α$
$1000 = 200 \times α$
$α = 5 rad / \sec^{2}$
From equation of motion,
$ω = ω_{0} + α t$
$ω = 0 + 5 \times 3$
$ω = 15 rad / \sec$

CPMT-1999

Rotational Motion

149923 The angular momentum of a particle with respect to the origin will not be zero, if

1 the directional line of linear momentum passes through the origin

2 the particle is at the origin (c) the angle between the position vector and linear momentum is

180^{\circ}

3 (d.) the linear momentum vanishes

4 the angle between the position vector and linear momentum is

90^{\circ}

Explanation:

E We know that,
angular momentum, $L = \vec{r} \times \vec{p} = rp \sin θ$
The conditions in which the value of angular momentum is zero, are given below.
(1) When the directional line of liner momentum passes through the origin it means the value of $θ$ will be zero so that $\sin θ = 0$ .
Hence angular momentum (L) will be zero
(2) When the value of position vector is zero, it means the particle is at the origin.
(3) When the value of $θ$ will be $180^{\circ}$ , it means the angle between the position vector and linear momentum is $180^{\circ}$ .
(4) When the value of linear momentum will be zero, it means the linear momentum vanishes.

Kerala CEE 04.07.2022

Rotational Motion

149924 A particle is moving in an elliptical orbit as shown in figure. If $\vec{p}, \vec{L}$ and $\vec{r}$ denote the linear momentum, angular momentum and position vector of the particle (from focus $O$ respectively at a point at $A$ , then the direction of $\vec{α} = \vec{p} \times \vec{L}$ is along.

1 + ve

x

axis

2 -ve

x

axis

3 + ve y axis

4 - ve y axis

Explanation:

A
From figure,
Linear momentum is moving anti-clockwise direction and perpendicular to radius from point A towards +ve $y$ -axis ( p $\hat{j}$ .
From right hand thumb Rule,
Angular momentum $\vec{L}$ is towards +ve z-axis $L \hat{k}$
Now,
$\vec{α} = \vec{p} \times \vec{L}$
$\vec{α} = (p \hat{j}) \times (L \hat{k}$
$= p L (\hat{j} \times \hat{k}$
$= p L \hat{i}$
From properties of cross product of 2 vector, therefore the direction of $α$ is +ve $x$ -axis.

WB JEE 2022

Rotational Motion

149925 If the radius of a spherical object, rotating about its diameter with a time period of 2 second, is reduced to half its actual value, keeping its mass unchanged, its time period becomes (assuming zero external torque)

1 Remains the same

2

6 s

3

0.5 s

4

1 s

Explanation:

C Initial angular momentum of spherical object
$L_{1} = I ω = \frac{2}{5} {MR}^{2} ω$
Final angular momentum of Spherical object
$L_{2} = I ω = \frac{2}{5} M {(\frac{R}{2})}^{2} ω^{'}$
$∵$ External torque is zero (Given)
So, Momentum will be conserved-
$L_{1} = L_{2}$
$\frac{2}{5} {MR}^{2} ω = \frac{2}{5} M {(\frac{R}{2})}^{2} ω^{'}$
$ω R^{2} = \frac{R^{2}}{4} ω^{'}$
$\Rightarrow \frac{1}{T} = \frac{1}{4 T^{'}}$
$\frac{1}{2} = \frac{1}{4 T^{'}} (T = 2 \sec .$
$T^{'} = \frac{1}{2} \sec .$
$T^{'} = 0.5 \sec$

AP EAMCET-03.09.2021

Rotational Motion

149922 A constant torque of $1000 N - m$ turns a wheel of moment of inertia $200 kg - m^{2}$ about an axis through its centre. Its angular velocity after 3 second is, in $rad / s$ :

1 1

2 5

3 15

4 10

Explanation:

C Given,
Torque $(τ) = 1000 N - m$
Moment of Inertia (I) $= 200 kg m^{2}$
Time $(t) = 3 \sec$
We know that,
$τ = I α$
$1000 = 200 \times α$
$α = 5 rad / \sec^{2}$
From equation of motion,
$ω = ω_{0} + α t$
$ω = 0 + 5 \times 3$
$ω = 15 rad / \sec$

CPMT-1999

Rotational Motion

149923 The angular momentum of a particle with respect to the origin will not be zero, if

1 the directional line of linear momentum passes through the origin

2 the particle is at the origin (c) the angle between the position vector and linear momentum is

180^{\circ}

3 (d.) the linear momentum vanishes

4 the angle between the position vector and linear momentum is

90^{\circ}

Explanation:

E We know that,
angular momentum, $L = \vec{r} \times \vec{p} = rp \sin θ$
The conditions in which the value of angular momentum is zero, are given below.
(1) When the directional line of liner momentum passes through the origin it means the value of $θ$ will be zero so that $\sin θ = 0$ .
Hence angular momentum (L) will be zero
(2) When the value of position vector is zero, it means the particle is at the origin.
(3) When the value of $θ$ will be $180^{\circ}$ , it means the angle between the position vector and linear momentum is $180^{\circ}$ .
(4) When the value of linear momentum will be zero, it means the linear momentum vanishes.

Kerala CEE 04.07.2022

Rotational Motion

149924 A particle is moving in an elliptical orbit as shown in figure. If $\vec{p}, \vec{L}$ and $\vec{r}$ denote the linear momentum, angular momentum and position vector of the particle (from focus $O$ respectively at a point at $A$ , then the direction of $\vec{α} = \vec{p} \times \vec{L}$ is along.

1 + ve

x

axis

2 -ve

x

axis

3 + ve y axis

4 - ve y axis

Explanation:

A
From figure,
Linear momentum is moving anti-clockwise direction and perpendicular to radius from point A towards +ve $y$ -axis ( p $\hat{j}$ .
From right hand thumb Rule,
Angular momentum $\vec{L}$ is towards +ve z-axis $L \hat{k}$
Now,
$\vec{α} = \vec{p} \times \vec{L}$
$\vec{α} = (p \hat{j}) \times (L \hat{k}$
$= p L (\hat{j} \times \hat{k}$
$= p L \hat{i}$
From properties of cross product of 2 vector, therefore the direction of $α$ is +ve $x$ -axis.

WB JEE 2022

Rotational Motion

149925 If the radius of a spherical object, rotating about its diameter with a time period of 2 second, is reduced to half its actual value, keeping its mass unchanged, its time period becomes (assuming zero external torque)

1 Remains the same

2

6 s

3

0.5 s

4

1 s

Explanation:

C Initial angular momentum of spherical object
$L_{1} = I ω = \frac{2}{5} {MR}^{2} ω$
Final angular momentum of Spherical object
$L_{2} = I ω = \frac{2}{5} M {(\frac{R}{2})}^{2} ω^{'}$
$∵$ External torque is zero (Given)
So, Momentum will be conserved-
$L_{1} = L_{2}$
$\frac{2}{5} {MR}^{2} ω = \frac{2}{5} M {(\frac{R}{2})}^{2} ω^{'}$
$ω R^{2} = \frac{R^{2}}{4} ω^{'}$
$\Rightarrow \frac{1}{T} = \frac{1}{4 T^{'}}$
$\frac{1}{2} = \frac{1}{4 T^{'}} (T = 2 \sec .$
$T^{'} = \frac{1}{2} \sec .$
$T^{'} = 0.5 \sec$

AP EAMCET-03.09.2021

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