Q: A constant torque acting on a uniform circular wheel changes its angular momentum from \$L\$ to \$4L\$ in \$4\$ seconds. The torque acting on it is:

As we know that, \$\tau=\dfrac{\Delta{L}}{{\Delta}t}\$So, \$\tau=\dfrac{4L-L}{4}=\dfrac{3L}{4}\$

Q: A particle of mass \$m\$ moves uniformly with a speed \$v\$ along a circle of radius \$r\$. A and B are two points on the circle, such that the arc \$AB\$ subtends an angle \$\theta\$ at the center of the circle. The magnitude of change in momentum as the particle moves from \$A\$ to \$B\$ is given by:

Assume particle is at lowest point of the circle\$P_i= mv\hat{i}\$After it rotates \$\theta\$ angle \$P_f= mv\cos \theta \hat{i}+mv\sin \theta \hat{j}\$Change in momentum \$=P_f-P_1 =\$ \$mv\cos \theta \hat{i}+mv\sin \theta \hat{j} -mv\hat{i}\$ \$ = mv(-1+\cos \theta) \hat{i}+mv\sin \theta \hat{j} \$\$ |\Delta P|= mv\sqrt{(-1+\cos \theta)^2+\sin^2 \theta \ }\$\$ |\Delta P|= mv\sqrt{1+\cos^2 \theta-2\cos \theta+\sin^2 \theta }\$\$ |\Delta P|= 2mv\sin (\dfrac{ \theta }{2}) \$

Q: The kinetic energy of a moving body is given by \$K = 2v^{2}\$, \$K\$ being in joules and \$v\$ in m/s. Its momentum when traveling with a velocity of \$2\ ms^{-1}\$ will be :

\$K=\dfrac{1}{2}mv^{2}=2v^{2}\$\$\Rightarrow m=4kg\$\$\vec{p}=m.\vec{v}\$ \$=4\times 2\$ \$=8\ kg m/s\$

Q: A torque of \$0.5 Nm\$ is required to drive a screw into a wooden frame with the help of a screw driver. If one of the two forces of couple produced by screw driver is \$50 N\$, the width of the screw driver is:

Let width be \$x\$. So the torque due to couple is \$2\times50\times x/2=50x\$\$50x = 0.5\Rightarrow x=0.01m=1\ cm\$

Q: The position of axis of rotation of a body is changed so that its moment of inertia decreases by 36%. Find the % change in its radius of gyration.

\$\textbf{Step 1 - Finding relation between}\$ \$k_{1}\ and\ k_{2}\$We know that \$I = mk^{2}\$ Initially, \$I_{1} = mk_{1}^{2}\$ and finally, \$I_{2} = mk_{2}^{2}\$given that initial moment of inertia decrease by \$36\%\$\$\therefore I_{2} = I_{1} - \dfrac {36}{100}I_{1}\$\$\Rightarrow I_{2} = 0.64 I_{1}\$\$\Rightarrow mk_{2}^{2} = 0.64\ mk_{1}^{2}\$\$\Rightarrow k_{2} = 0.8 k_{1}\$, so \$k_{1} > k_{2}\$\$\textbf{Step 2 - Calculating change in radius of gyration}\$Change in radius of gyration \$= \dfrac {k_{1} - k_{2}}{k_{1}}\times 100\$ \$= \dfrac {k_{1} - 0.8k_{1}}{k_{1}} \times 100\$ \$= 20\$Hence, the radius of gyration decreases by \$20\%\$

Q: A wheel of radius \$R\$ is free to rotate about its natural axis and \$I\$ is its moment of inertia. If a tangential force \$F\$ is applied on the wheel along its rim, then angular acceleration of wheel is:

The torque is given using the relation \$\vec{\tau} = \vec{F} \times \vec{R}\$or\$\tau=I\alpha\$or\$\alpha=\dfrac{FR}{I}\$

Q: A flywheel making \$120\$ r.p.m is acted upon by a retarding torque producing angular retardation of \$\pi \ rad/s^{2}\$. Time taken by it to come to rest is:

\$T=I \alpha \$\$\omega =\dfrac{120\times 2\pi }{60}\$\$\omega =4\pi \$\$\omega =\omega _{0}+\alpha t\$\$0=4\pi -\pi t\$\$t=4 s\$

Q: A brass disc is rotating about its axis. If temperature of the disc is increased then its:

Radius of gyration andmoment of inertia will increase.\$\because I=mk^{2}\$\$k=\sqrt{\dfrac{I}{m}}\$\$I=\dfrac{mR^{2}}{2}\$\$k=\dfrac{R}{\sqrt{2}}\$\$A=A_{0}(1+ 2\alpha \Delta T) (eq^{n} \$ of thermal expansion)\$\pi r^{2}=\pi R_{0}^{2} (1+ 2\alpha \Delta T)\$\$r^{2}=R_{0}^{2} (1+ 2\alpha \Delta T)\$\$\therefore\$ r will increase on increasing the temperature\$\therefore I\ and \ k\$ will increase on increasing the \$temperature\$

Q: Four particles each of mass \$m\$ are placed at the corners of a square of side length \$l\$. The radius of gyration of the system about an axis perpendicular to the square and passing through its center is:

The distance of mass from the center of square is \$r=\dfrac{l}{\sqrt{2}}\$So the total moment of inertia of system \$I=4mr^2=2ml^2\$ Let the radius of gyration be k\$I=2ml^2=4mk^2\Rightarrow k=\dfrac{l}{\sqrt{2}}\$

Question 1

A constant torque acting on a uniform circular wheel changes its angular momentum from $L$ to $4L$ in $4$ seconds. The torque acting on it is:

Accepted Answer

As we know that, $\tau=\dfrac{\Delta{L}}{{\Delta}t}$So, $\tau=\dfrac{4L-L}{4}=\dfrac{3L}{4}$

Question 2

A particle of mass $m$ moves uniformly with a speed $v$ along a circle of radius $r$. A and B are two points on the circle, such that the arc $AB$ subtends an angle $\theta$ at the center of the circle. The magnitude of change in momentum as the particle moves from $A$ to $B$ is given by:

Accepted Answer

Assume particle is at lowest point of the circle$P_i= mv\hat{i}$After it rotates $\theta$ angle $P_f= mv\cos \theta \hat{i}+mv\sin \theta \hat{j}$Change in momentum $=P_f-P_1 =$ $mv\cos \theta \hat{i}+mv\sin \theta \hat{j} -mv\hat{i}$                                      $ = mv(-1+\cos \theta) \hat{i}+mv\sin \theta \hat{j} $$ |\Delta P|= mv\sqrt{(-1+\cos \theta)^2+\sin^2 \theta \ }$$ |\Delta P|= mv\sqrt{1+\cos^2 \theta-2\cos \theta+\sin^2 \theta }$$ |\Delta P|= 2mv\sin (\dfrac{ \theta }{2}) $

Question 3

The kinetic energy of a moving body is given by $K = 2v^{2}$, $K$ being in joules and $v$ in m/s. Its momentum when traveling with a velocity of $2\ ms^{-1}$ will be :

Accepted Answer

$K=\dfrac{1}{2}mv^{2}=2v^{2}$$\Rightarrow m=4kg$$\vec{p}=m.\vec{v}$    $=4\times 2$    $=8\ kg m/s$

Question 4

A torque of $0.5  Nm$ is required to drive a screw into a wooden frame with the help of a screw driver. If one of the two forces of couple produced by screw driver is $50  N$, the width of the screw driver is:

Accepted Answer

Let width be $x$. So the torque due to couple is $2\times50\times x/2=50x$$50x = 0.5\Rightarrow x=0.01m=1\ cm$

Question 5

The position of axis of rotation of a body is changed so that its moment of inertia decreases by 36%. Find the % change in its radius of gyration.

Accepted Answer

$\textbf{Step 1 - Finding relation between}$ $k_{1}\ and\ k_{2}$We know  that $I = mk^{2}$          Initially, $I_{1} = mk_{1}^{2}$     and finally, $I_{2} = mk_{2}^{2}$given that initial moment of inertia decrease by $36\%$$\therefore I_{2} = I_{1} - \dfrac {36}{100}I_{1}$$\Rightarrow I_{2} = 0.64 I_{1}$$\Rightarrow mk_{2}^{2} = 0.64\ mk_{1}^{2}$$\Rightarrow k_{2} = 0.8 k_{1}$, so $k_{1} > k_{2}$$\textbf{Step 2 - Calculating change in radius of gyration}$Change in radius of gyration $= \dfrac {k_{1} - k_{2}}{k_{1}}\times 100$                                                 $= \dfrac {k_{1} - 0.8k_{1}}{k_{1}} \times 100$                                                 $= 20$Hence, the radius of gyration decreases by $20\%$

Question 6

The radius of gyration of a rotating metallic disc is independent of which of the following physical quantity.

Accepted Answer

Toppr Ask Answer

Question 7

A wheel of radius $R$ is free to rotate about its natural axis and $I$ is its moment of inertia. If a tangential force $F$ is applied on the wheel along its rim, then angular acceleration of wheel is:

Accepted Answer

The torque is given using the relation $\vec{\tau} = \vec{F} \times \vec{R}$or$\tau=I\alpha$or$\alpha=\dfrac{FR}{I}$

Question 8

A flywheel making $120$ r.p.m is acted upon by a retarding torque producing angular retardation of $\pi  \ rad/s^{2}$. Time taken by it to come to rest is:

Accepted Answer

$T=I  \alpha $$\omega =\dfrac{120\times 2\pi }{60}$$\omega =4\pi $$\omega =\omega _{0}+\alpha   t$$0=4\pi -\pi t$$t=4 s$

Question 9

A brass disc is rotating about its axis. If temperature of the disc is increased then its:

Accepted Answer

Radius of gyration andmoment of inertia will increase.$\because I=mk^{2}$$k=\sqrt{\dfrac{I}{m}}$$I=\dfrac{mR^{2}}{2}$$k=\dfrac{R}{\sqrt{2}}$$A=A_{0}(1+ 2\alpha  \Delta T)  (eq^{n} $ of thermal expansion)$\pi r^{2}=\pi R_{0}^{2} (1+ 2\alpha \Delta T)$$r^{2}=R_{0}^{2} (1+ 2\alpha  \Delta T)$$\therefore$ r will increase on increasing the temperature$\therefore I\  and \ k$ will increase on increasing the $temperature$

Question 10

Four particles each of mass $m$ are placed at the corners of a square of side length $l$ . The radius of gyration of the system about an axis perpendicular to the square and passing through its center is:

Accepted Answer

The distance of mass from the center of square is $r=\dfrac{l}{\sqrt{2}}$So the total moment of inertia of system $I=4mr^2=2ml^2$ Let the radius of gyration be k$I=2ml^2=4mk^2\Rightarrow k=\dfrac{l}{\sqrt{2}}$

Systems Of Particles And Rotational Motion

Physics
6914 Views

A constant torque acting on a uniform circular wheel changes its angular momentum from $L$ to $4 L$ in $4$ seconds. The torque acting on it is:

A particle of mass $m$ moves uniformly with a speed $v$ along a circle of radius $r$ . A and B are two points on the circle, such that the arc $A B$ subtends an angle $θ$ at the center of the circle. The magnitude of change in momentum as the particle moves from $A$ to $B$ is given by:

The kinetic energy of a moving body is given by $K = 2 v^{2}$ , $K$ being in joules and $v$ in m/s. Its momentum when traveling with a velocity of $2 m s^{- 1}$ will be :

A torque of $0.5 N m$ is required to drive a screw into a wooden frame with the help of a screw driver. If one of the two forces of couple produced by screw driver is $50 N$ , the width of the screw driver is:

The position of axis of rotation of a body is changed so that its moment of inertia decreases by 36%. Find the % change in its radius of gyration.

The radius of gyration of a rotating metallic disc is independent of which of the following physical quantity.

A wheel of radius $R$ is free to rotate about its natural axis and $I$ is its moment of inertia. If a tangential force $F$ is applied on the wheel along its rim, then angular acceleration of wheel is:

A flywheel making $120$ r.p.m is acted upon by a retarding torque producing angular retardation of $π r a d / s^{2}$ . Time taken by it to come to rest is:

A brass disc is rotating about its axis. If temperature of the disc is increased then its:

Four particles each of mass $m$ are placed at the corners of a square of side length $l$ . The radius of gyration of the system about an axis perpendicular to the square and passing through its center is:

Systems Of Particles And Rotational Motion

Physics 6914 Views

A constant torque acting on a uniform circular wheel changes its angular momentum from L to 4L in 4 seconds. The torque acting on it is:

A particle of mass m moves uniformly with a speed v along a circle of radius r. A and B are two points on the circle, such that the arc AB subtends an angle θ at the center of the circle. The magnitude of change in momentum as the particle moves from A to B is given by:

The kinetic energy of a moving body is given by K=2v2, K being in joules and v in m/s. Its momentum when traveling with a velocity of 2 ms−1 will be :

A torque of 0.5Nm is required to drive a screw into a wooden frame with the help of a screw driver. If one of the two forces of couple produced by screw driver is 50N, the width of the screw driver is:

The position of axis of rotation of a body is changed so that its moment of inertia decreases by 36%. Find the % change in its radius of gyration.

The radius of gyration of a rotating metallic disc is independent of which of the following physical quantity.

A wheel of radius R is free to rotate about its natural axis and I is its moment of inertia. If a tangential force F is applied on the wheel along its rim, then angular acceleration of wheel is:

A flywheel making 120 r.p.m is acted upon by a retarding torque producing angular retardation of π rad/s2. Time taken by it to come to rest is:

A brass disc is rotating about its axis. If temperature of the disc is increased then its:

Four particles each of mass m are placed at the corners of a square of side length l. The radius of gyration of the system about an axis perpendicular to the square and passing through its center is:

Physics
6914 Views

A constant torque acting on a uniform circular wheel changes its angular momentum from $L$ to $4 L$ in $4$ seconds. The torque acting on it is:

A particle of mass $m$ moves uniformly with a speed $v$ along a circle of radius $r$ . A and B are two points on the circle, such that the arc $A B$ subtends an angle $θ$ at the center of the circle. The magnitude of change in momentum as the particle moves from $A$ to $B$ is given by:

The kinetic energy of a moving body is given by $K = 2 v^{2}$ , $K$ being in joules and $v$ in m/s. Its momentum when traveling with a velocity of $2 m s^{- 1}$ will be :

A torque of $0.5 N m$ is required to drive a screw into a wooden frame with the help of a screw driver. If one of the two forces of couple produced by screw driver is $50 N$ , the width of the screw driver is:

A wheel of radius $R$ is free to rotate about its natural axis and $I$ is its moment of inertia. If a tangential force $F$ is applied on the wheel along its rim, then angular acceleration of wheel is:

A flywheel making $120$ r.p.m is acted upon by a retarding torque producing angular retardation of $π r a d / s^{2}$ . Time taken by it to come to rest is:

Four particles each of mass $m$ are placed at the corners of a square of side length $l$ . The radius of gyration of the system about an axis perpendicular to the square and passing through its center is: