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Rotational Motion

149937 When the torque acting on a system is zero, which of the following remains constant?

1 linear momentum

2 angular momentum

3 linear impulse

4 linear velocity

Explanation:

B When no external torque acting on the system then angular momentum of a system of particle remains constant.
$\frac{dL}{dt} = τ$
$∵ τ = 0$
$\frac{dL}{dt} = 0$
$∴ L =$ constant
Hence, Total angular momentum of the system is remains constant.

AP EAMCET-25.09.2020

Rotational Motion

149938 The moment of inertia of a flywheel making 300 revolutions per minute is $0.3 kg m^{2}$ . Find the torque required to bring it to rest in 20 seconds.

1 47.1 N.m

2

- 47.1 N . m

3 -0.471 N.m

4

0.471 N . m

Explanation:

C Given,
Moment of Inertia (I) $= 0.3 kg m^{2}$
Time $t) = 20 s$
Rotation per minute $= 300 rpm$
$f = \frac{300}{60} = 5$ Rotation per sec
$ω_{o} = 2 π f = 2 π \times 5 = 10 π rad / \sec$
We know that,
$ω = ω_{0} + α t$
$ω = ω_{0} - α t$
Where, final angular acceleration $ω) = 0$
$ω_{0} = α t$
$10 π = α \times 20$
$α = \frac{10 π}{20} = \frac{π}{2}$
Torque $τ) = I α = 0.3 \times \frac{π}{2} = 0.15 π$
$= 0.15 \times 3.14$
$τ = 0.471 Nm$
Torque is opposes the motion of fly wheel.
So, most correct answer is (c).
$τ = - 0.471 Nm$

AP EAMCET-24.09.2020

Rotational Motion

149939 A force of $(7 \hat{i} + 3 \hat{j} - 5 \hat{k}) N$ acts on a particle whose position vector is $\hat{i} - \hat{j} + \hat{k}) m$ . Find the torque experienced by the particle about the origin in $N - m$ .

1

2 \hat{i} + 12 \hat{j} + 10 \hat{k}

2

3 \hat{i} + 10 \hat{j} + 12 \hat{k}

3

10 \hat{i} + 12 \hat{j} + 2 \hat{k}

4

12 \hat{i} + 10 \hat{j} + 2 \hat{k}

Explanation:

A Force, $F = (7 \hat{i} + 3 \hat{j} - 5 \hat{k}) N$
Distance, $r = (\hat{i} - \hat{j} + \hat{k}) m$
Torque will be
$\vec{τ} = \vec{F} \times \vec{r}$
$τ = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 1 & + 1 \\ 7 & 3 & - 5 \end{array} |$
$= \hat{i} (5 - 3) - \hat{j} (- 5 - 7) + \hat{k} (3 + 7$
$= 2 \hat{i} + 12 \hat{j} + 10 \hat{k}$

TS EAMCET 29.09.2020

Rotational Motion

149941 A constant torque of $200 N - m$ turns a flywheel, which is at rest, about an axis through its centre and perpendicular to its plane. If its moment of inertia is $50 kg - m^{2}$ , then in 4 second, what will be change in its angular momentum?

1

200 kg - m^{2} / s

2

800 kg - m^{2} / s

3

20 kg - m^{2} / s

4

40 kg - m^{2} / s

Explanation:

B Given,
$τ = 200 Nm,$
$t = 4 \sec,$
$I = 50 kg m^{2}$
We know,
Change in angular momentum $=$ Torque $\times$ time
$= τ \times t$
$= 200 \times 4$
$= 800 kg m^{2} / s$

MHT-CET 2020

Rotational Motion

149937 When the torque acting on a system is zero, which of the following remains constant?

1 linear momentum

2 angular momentum

3 linear impulse

4 linear velocity

Explanation:

B When no external torque acting on the system then angular momentum of a system of particle remains constant.
$\frac{dL}{dt} = τ$
$∵ τ = 0$
$\frac{dL}{dt} = 0$
$∴ L =$ constant
Hence, Total angular momentum of the system is remains constant.

AP EAMCET-25.09.2020

Rotational Motion

149938 The moment of inertia of a flywheel making 300 revolutions per minute is $0.3 kg m^{2}$ . Find the torque required to bring it to rest in 20 seconds.

1 47.1 N.m

2

- 47.1 N . m

3 -0.471 N.m

4

0.471 N . m

Explanation:

C Given,
Moment of Inertia (I) $= 0.3 kg m^{2}$
Time $t) = 20 s$
Rotation per minute $= 300 rpm$
$f = \frac{300}{60} = 5$ Rotation per sec
$ω_{o} = 2 π f = 2 π \times 5 = 10 π rad / \sec$
We know that,
$ω = ω_{0} + α t$
$ω = ω_{0} - α t$
Where, final angular acceleration $ω) = 0$
$ω_{0} = α t$
$10 π = α \times 20$
$α = \frac{10 π}{20} = \frac{π}{2}$
Torque $τ) = I α = 0.3 \times \frac{π}{2} = 0.15 π$
$= 0.15 \times 3.14$
$τ = 0.471 Nm$
Torque is opposes the motion of fly wheel.
So, most correct answer is (c).
$τ = - 0.471 Nm$

AP EAMCET-24.09.2020

Rotational Motion

149939 A force of $(7 \hat{i} + 3 \hat{j} - 5 \hat{k}) N$ acts on a particle whose position vector is $\hat{i} - \hat{j} + \hat{k}) m$ . Find the torque experienced by the particle about the origin in $N - m$ .

1

2 \hat{i} + 12 \hat{j} + 10 \hat{k}

2

3 \hat{i} + 10 \hat{j} + 12 \hat{k}

3

10 \hat{i} + 12 \hat{j} + 2 \hat{k}

4

12 \hat{i} + 10 \hat{j} + 2 \hat{k}

Explanation:

A Force, $F = (7 \hat{i} + 3 \hat{j} - 5 \hat{k}) N$
Distance, $r = (\hat{i} - \hat{j} + \hat{k}) m$
Torque will be
$\vec{τ} = \vec{F} \times \vec{r}$
$τ = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 1 & + 1 \\ 7 & 3 & - 5 \end{array} |$
$= \hat{i} (5 - 3) - \hat{j} (- 5 - 7) + \hat{k} (3 + 7$
$= 2 \hat{i} + 12 \hat{j} + 10 \hat{k}$

TS EAMCET 29.09.2020

Rotational Motion

149941 A constant torque of $200 N - m$ turns a flywheel, which is at rest, about an axis through its centre and perpendicular to its plane. If its moment of inertia is $50 kg - m^{2}$ , then in 4 second, what will be change in its angular momentum?

1

200 kg - m^{2} / s

2

800 kg - m^{2} / s

3

20 kg - m^{2} / s

4

40 kg - m^{2} / s

Explanation:

B Given,
$τ = 200 Nm,$
$t = 4 \sec,$
$I = 50 kg m^{2}$
We know,
Change in angular momentum $=$ Torque $\times$ time
$= τ \times t$
$= 200 \times 4$
$= 800 kg m^{2} / s$

MHT-CET 2020

Rotational Motion

149937 When the torque acting on a system is zero, which of the following remains constant?

1 linear momentum

2 angular momentum

3 linear impulse

4 linear velocity

Explanation:

B When no external torque acting on the system then angular momentum of a system of particle remains constant.
$\frac{dL}{dt} = τ$
$∵ τ = 0$
$\frac{dL}{dt} = 0$
$∴ L =$ constant
Hence, Total angular momentum of the system is remains constant.

AP EAMCET-25.09.2020

Rotational Motion

149938 The moment of inertia of a flywheel making 300 revolutions per minute is $0.3 kg m^{2}$ . Find the torque required to bring it to rest in 20 seconds.

1 47.1 N.m

2

- 47.1 N . m

3 -0.471 N.m

4

0.471 N . m

Explanation:

C Given,
Moment of Inertia (I) $= 0.3 kg m^{2}$
Time $t) = 20 s$
Rotation per minute $= 300 rpm$
$f = \frac{300}{60} = 5$ Rotation per sec
$ω_{o} = 2 π f = 2 π \times 5 = 10 π rad / \sec$
We know that,
$ω = ω_{0} + α t$
$ω = ω_{0} - α t$
Where, final angular acceleration $ω) = 0$
$ω_{0} = α t$
$10 π = α \times 20$
$α = \frac{10 π}{20} = \frac{π}{2}$
Torque $τ) = I α = 0.3 \times \frac{π}{2} = 0.15 π$
$= 0.15 \times 3.14$
$τ = 0.471 Nm$
Torque is opposes the motion of fly wheel.
So, most correct answer is (c).
$τ = - 0.471 Nm$

AP EAMCET-24.09.2020

Rotational Motion

149939 A force of $(7 \hat{i} + 3 \hat{j} - 5 \hat{k}) N$ acts on a particle whose position vector is $\hat{i} - \hat{j} + \hat{k}) m$ . Find the torque experienced by the particle about the origin in $N - m$ .

1

2 \hat{i} + 12 \hat{j} + 10 \hat{k}

2

3 \hat{i} + 10 \hat{j} + 12 \hat{k}

3

10 \hat{i} + 12 \hat{j} + 2 \hat{k}

4

12 \hat{i} + 10 \hat{j} + 2 \hat{k}

Explanation:

A Force, $F = (7 \hat{i} + 3 \hat{j} - 5 \hat{k}) N$
Distance, $r = (\hat{i} - \hat{j} + \hat{k}) m$
Torque will be
$\vec{τ} = \vec{F} \times \vec{r}$
$τ = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 1 & + 1 \\ 7 & 3 & - 5 \end{array} |$
$= \hat{i} (5 - 3) - \hat{j} (- 5 - 7) + \hat{k} (3 + 7$
$= 2 \hat{i} + 12 \hat{j} + 10 \hat{k}$

TS EAMCET 29.09.2020

Rotational Motion

149941 A constant torque of $200 N - m$ turns a flywheel, which is at rest, about an axis through its centre and perpendicular to its plane. If its moment of inertia is $50 kg - m^{2}$ , then in 4 second, what will be change in its angular momentum?

1

200 kg - m^{2} / s

2

800 kg - m^{2} / s

3

20 kg - m^{2} / s

4

40 kg - m^{2} / s

Explanation:

B Given,
$τ = 200 Nm,$
$t = 4 \sec,$
$I = 50 kg m^{2}$
We know,
Change in angular momentum $=$ Torque $\times$ time
$= τ \times t$
$= 200 \times 4$
$= 800 kg m^{2} / s$

MHT-CET 2020

Rotational Motion

149937 When the torque acting on a system is zero, which of the following remains constant?

1 linear momentum

2 angular momentum

3 linear impulse

4 linear velocity

Explanation:

B When no external torque acting on the system then angular momentum of a system of particle remains constant.
$\frac{dL}{dt} = τ$
$∵ τ = 0$
$\frac{dL}{dt} = 0$
$∴ L =$ constant
Hence, Total angular momentum of the system is remains constant.

AP EAMCET-25.09.2020

Rotational Motion

149938 The moment of inertia of a flywheel making 300 revolutions per minute is $0.3 kg m^{2}$ . Find the torque required to bring it to rest in 20 seconds.

1 47.1 N.m

2

- 47.1 N . m

3 -0.471 N.m

4

0.471 N . m

Explanation:

C Given,
Moment of Inertia (I) $= 0.3 kg m^{2}$
Time $t) = 20 s$
Rotation per minute $= 300 rpm$
$f = \frac{300}{60} = 5$ Rotation per sec
$ω_{o} = 2 π f = 2 π \times 5 = 10 π rad / \sec$
We know that,
$ω = ω_{0} + α t$
$ω = ω_{0} - α t$
Where, final angular acceleration $ω) = 0$
$ω_{0} = α t$
$10 π = α \times 20$
$α = \frac{10 π}{20} = \frac{π}{2}$
Torque $τ) = I α = 0.3 \times \frac{π}{2} = 0.15 π$
$= 0.15 \times 3.14$
$τ = 0.471 Nm$
Torque is opposes the motion of fly wheel.
So, most correct answer is (c).
$τ = - 0.471 Nm$

AP EAMCET-24.09.2020

Rotational Motion

149939 A force of $(7 \hat{i} + 3 \hat{j} - 5 \hat{k}) N$ acts on a particle whose position vector is $\hat{i} - \hat{j} + \hat{k}) m$ . Find the torque experienced by the particle about the origin in $N - m$ .

1

2 \hat{i} + 12 \hat{j} + 10 \hat{k}

2

3 \hat{i} + 10 \hat{j} + 12 \hat{k}

3

10 \hat{i} + 12 \hat{j} + 2 \hat{k}

4

12 \hat{i} + 10 \hat{j} + 2 \hat{k}

Explanation:

A Force, $F = (7 \hat{i} + 3 \hat{j} - 5 \hat{k}) N$
Distance, $r = (\hat{i} - \hat{j} + \hat{k}) m$
Torque will be
$\vec{τ} = \vec{F} \times \vec{r}$
$τ = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 1 & + 1 \\ 7 & 3 & - 5 \end{array} |$
$= \hat{i} (5 - 3) - \hat{j} (- 5 - 7) + \hat{k} (3 + 7$
$= 2 \hat{i} + 12 \hat{j} + 10 \hat{k}$

TS EAMCET 29.09.2020

Rotational Motion

149941 A constant torque of $200 N - m$ turns a flywheel, which is at rest, about an axis through its centre and perpendicular to its plane. If its moment of inertia is $50 kg - m^{2}$ , then in 4 second, what will be change in its angular momentum?

1

200 kg - m^{2} / s

2

800 kg - m^{2} / s

3

20 kg - m^{2} / s

4

40 kg - m^{2} / s

Explanation:

B Given,
$τ = 200 Nm,$
$t = 4 \sec,$
$I = 50 kg m^{2}$
We know,
Change in angular momentum $=$ Torque $\times$ time
$= τ \times t$
$= 200 \times 4$
$= 800 kg m^{2} / s$

MHT-CET 2020

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