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PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366032 A loop and a disc roll without slipping with same linear velocity $v$ . The mass of the loop and the disc is same. If the total kinetic energy of the loop is 8 $J$ , find the kinetic energy of the disc

1

3 J

2

6 J

3

9 J

4

1 J

Explanation:

Total kinetic energy of loop
$= \frac{1}{2} I ω + \frac{1}{2} M v^{2}$ $= \frac{1}{2} (M R^{2} ω^{2}) + \frac{1}{2} M v^{2} = M v^{2} = 8 J (given)$
Total kinetic energy of disc
$= \frac{1}{2} I ω^{2} + \frac{1}{2} M v^{2}$
$= \frac{1}{2} (\frac{1}{2} M R^{2}) (\frac{v^{2}}{R^{2}}) + \frac{1}{2} M v^{2}$
$= \frac{3}{4} M v^{2} = \frac{3}{4} (8) = 6 J$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366033 A body is rolling down an inclined plane. If $K E$ of rotation is $40 %$ $K . E .$ of in translatory state, then the body is a

1 ring

2 cylinder

3 hollow ball

4 solid ball

Explanation:

$K E$ of rotation is $40 %$ of in translatory
$\begin{aligned} (K E)_{rot} & = 40 % \times (K E)_{trans} \\ \frac{1}{2} I ω^{2} & = \frac{40}{100} \times \frac{1}{2} m v^{2} \\ \Rightarrow I ω^{2} & = \frac{2}{5} m (ω r)^{2} [∵ v = ω r] \\ \Rightarrow I & = \frac{2}{5} m r^{2} \end{aligned}$
where, $I$ is moment of inertia.Thus, the body is a solid sphere (solid ball).

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366034 A ball rolls without slipping. The radius of gyration of the ball about an axis passing through its centre of mass is $K$ . If radius of the ball be $R$ , then the fraction of total enery associated with its rotational energy will be

1

\frac{K^{2}}{R^{2}}

2

\frac{K^{2}}{K^{2} + R^{2}}

3

\frac{R^{2}}{K^{2} + R^{2}}

4

\frac{K^{2} + R^{2}}{R^{2}}

Explanation:

Denote Kinetic energy $= E_{k}$
$\frac{{(E_{k})}_{r o t}}{E_{T}} = \frac{\frac{1}{2} m v^{2} (\frac{K^{2}}{R^{2}})}{\frac{1}{2} m v^{2} (1 + \frac{K^{2}}{R^{2}})} = \frac{K^{2}}{K^{2} + R^{2}}$
Where subscript (rot) is for rotational and subscript (T) for total. So, option (2) is correct.

MHTCET - 2021

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366035 A body is rolling without slipping on a horizontal surface and its rotational kinetic energy is equal to the translational kinetic energy. The body is

1 Sphere

2 Ring

3 Disc

4 Cylinder

Explanation:

$K_{R} = K_{T} \Rightarrow \frac{1}{2} m v^{2} (\frac{K^{2}}{R^{2}}) = \frac{1}{2} m v^{2} \Rightarrow \frac{K^{2}}{R^{2}} = 1$
i.e., the body is ring

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366036 A uniform disc of mass $m$ is fitted (pivoted smoothly) with a rod of mass $\frac{m}{2}$ . If the bottom of the rod is pulled with a velocity $v$ it moves without changing its angle of orientation and the disc rolls without sliding. The kinetic energy of the system(rod+disc) is
supporting img

1

\frac{m v^{2}}{4}

2

\frac{m v^{2}}{4}

3

\frac{15 m v^{2}}{16}

4

m v^{2}

Explanation:

Rolling can be considered as pure rotation about point of contact
$K_{r o l l i n g} = \frac{1}{2} I_{P} ω^{2}$
$= \frac{1}{2} (\frac{m R^{2}}{2} + m R^{2}) ω^{2}$
$= \frac{3}{4} m R^{2} ω^{2} (1)$
The rod translates with the velocity $v$ hence velocity of centre of disc will also be $v$
$v = ω R (2)$
From (1) and (2); $K_{rolling} = \frac{3}{4} m v^{2}$
Kinetic energy of the rod
$\begin{aligned} K_{rod} = \frac{1}{2} (\frac{m}{2}) v^{2} = \frac{m v^{2}}{4} \\ K_{total} = k_{rolling} + k_{rod} = m v^{2} \end{aligned}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366032 A loop and a disc roll without slipping with same linear velocity $v$ . The mass of the loop and the disc is same. If the total kinetic energy of the loop is 8 $J$ , find the kinetic energy of the disc

1

3 J

2

6 J

3

9 J

4

1 J

Explanation:

Total kinetic energy of loop
$= \frac{1}{2} I ω + \frac{1}{2} M v^{2}$ $= \frac{1}{2} (M R^{2} ω^{2}) + \frac{1}{2} M v^{2} = M v^{2} = 8 J (given)$
Total kinetic energy of disc
$= \frac{1}{2} I ω^{2} + \frac{1}{2} M v^{2}$
$= \frac{1}{2} (\frac{1}{2} M R^{2}) (\frac{v^{2}}{R^{2}}) + \frac{1}{2} M v^{2}$
$= \frac{3}{4} M v^{2} = \frac{3}{4} (8) = 6 J$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366033 A body is rolling down an inclined plane. If $K E$ of rotation is $40 %$ $K . E .$ of in translatory state, then the body is a

1 ring

2 cylinder

3 hollow ball

4 solid ball

Explanation:

$K E$ of rotation is $40 %$ of in translatory
$\begin{aligned} (K E)_{rot} & = 40 % \times (K E)_{trans} \\ \frac{1}{2} I ω^{2} & = \frac{40}{100} \times \frac{1}{2} m v^{2} \\ \Rightarrow I ω^{2} & = \frac{2}{5} m (ω r)^{2} [∵ v = ω r] \\ \Rightarrow I & = \frac{2}{5} m r^{2} \end{aligned}$
where, $I$ is moment of inertia.Thus, the body is a solid sphere (solid ball).

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366034 A ball rolls without slipping. The radius of gyration of the ball about an axis passing through its centre of mass is $K$ . If radius of the ball be $R$ , then the fraction of total enery associated with its rotational energy will be

1

\frac{K^{2}}{R^{2}}

2

\frac{K^{2}}{K^{2} + R^{2}}

3

\frac{R^{2}}{K^{2} + R^{2}}

4

\frac{K^{2} + R^{2}}{R^{2}}

Explanation:

Denote Kinetic energy $= E_{k}$
$\frac{{(E_{k})}_{r o t}}{E_{T}} = \frac{\frac{1}{2} m v^{2} (\frac{K^{2}}{R^{2}})}{\frac{1}{2} m v^{2} (1 + \frac{K^{2}}{R^{2}})} = \frac{K^{2}}{K^{2} + R^{2}}$
Where subscript (rot) is for rotational and subscript (T) for total. So, option (2) is correct.

MHTCET - 2021

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366035 A body is rolling without slipping on a horizontal surface and its rotational kinetic energy is equal to the translational kinetic energy. The body is

1 Sphere

2 Ring

3 Disc

4 Cylinder

Explanation:

$K_{R} = K_{T} \Rightarrow \frac{1}{2} m v^{2} (\frac{K^{2}}{R^{2}}) = \frac{1}{2} m v^{2} \Rightarrow \frac{K^{2}}{R^{2}} = 1$
i.e., the body is ring

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366036 A uniform disc of mass $m$ is fitted (pivoted smoothly) with a rod of mass $\frac{m}{2}$ . If the bottom of the rod is pulled with a velocity $v$ it moves without changing its angle of orientation and the disc rolls without sliding. The kinetic energy of the system(rod+disc) is
supporting img

1

\frac{m v^{2}}{4}

2

\frac{m v^{2}}{4}

3

\frac{15 m v^{2}}{16}

4

m v^{2}

Explanation:

Rolling can be considered as pure rotation about point of contact
$K_{r o l l i n g} = \frac{1}{2} I_{P} ω^{2}$
$= \frac{1}{2} (\frac{m R^{2}}{2} + m R^{2}) ω^{2}$
$= \frac{3}{4} m R^{2} ω^{2} (1)$
The rod translates with the velocity $v$ hence velocity of centre of disc will also be $v$
$v = ω R (2)$
From (1) and (2); $K_{rolling} = \frac{3}{4} m v^{2}$
Kinetic energy of the rod
$\begin{aligned} K_{rod} = \frac{1}{2} (\frac{m}{2}) v^{2} = \frac{m v^{2}}{4} \\ K_{total} = k_{rolling} + k_{rod} = m v^{2} \end{aligned}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366032 A loop and a disc roll without slipping with same linear velocity $v$ . The mass of the loop and the disc is same. If the total kinetic energy of the loop is 8 $J$ , find the kinetic energy of the disc

1

3 J

2

6 J

3

9 J

4

1 J

Explanation:

Total kinetic energy of loop
$= \frac{1}{2} I ω + \frac{1}{2} M v^{2}$ $= \frac{1}{2} (M R^{2} ω^{2}) + \frac{1}{2} M v^{2} = M v^{2} = 8 J (given)$
Total kinetic energy of disc
$= \frac{1}{2} I ω^{2} + \frac{1}{2} M v^{2}$
$= \frac{1}{2} (\frac{1}{2} M R^{2}) (\frac{v^{2}}{R^{2}}) + \frac{1}{2} M v^{2}$
$= \frac{3}{4} M v^{2} = \frac{3}{4} (8) = 6 J$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366033 A body is rolling down an inclined plane. If $K E$ of rotation is $40 %$ $K . E .$ of in translatory state, then the body is a

1 ring

2 cylinder

3 hollow ball

4 solid ball

Explanation:

$K E$ of rotation is $40 %$ of in translatory
$\begin{aligned} (K E)_{rot} & = 40 % \times (K E)_{trans} \\ \frac{1}{2} I ω^{2} & = \frac{40}{100} \times \frac{1}{2} m v^{2} \\ \Rightarrow I ω^{2} & = \frac{2}{5} m (ω r)^{2} [∵ v = ω r] \\ \Rightarrow I & = \frac{2}{5} m r^{2} \end{aligned}$
where, $I$ is moment of inertia.Thus, the body is a solid sphere (solid ball).

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366034 A ball rolls without slipping. The radius of gyration of the ball about an axis passing through its centre of mass is $K$ . If radius of the ball be $R$ , then the fraction of total enery associated with its rotational energy will be

1

\frac{K^{2}}{R^{2}}

2

\frac{K^{2}}{K^{2} + R^{2}}

3

\frac{R^{2}}{K^{2} + R^{2}}

4

\frac{K^{2} + R^{2}}{R^{2}}

Explanation:

Denote Kinetic energy $= E_{k}$
$\frac{{(E_{k})}_{r o t}}{E_{T}} = \frac{\frac{1}{2} m v^{2} (\frac{K^{2}}{R^{2}})}{\frac{1}{2} m v^{2} (1 + \frac{K^{2}}{R^{2}})} = \frac{K^{2}}{K^{2} + R^{2}}$
Where subscript (rot) is for rotational and subscript (T) for total. So, option (2) is correct.

MHTCET - 2021

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366035 A body is rolling without slipping on a horizontal surface and its rotational kinetic energy is equal to the translational kinetic energy. The body is

1 Sphere

2 Ring

3 Disc

4 Cylinder

Explanation:

$K_{R} = K_{T} \Rightarrow \frac{1}{2} m v^{2} (\frac{K^{2}}{R^{2}}) = \frac{1}{2} m v^{2} \Rightarrow \frac{K^{2}}{R^{2}} = 1$
i.e., the body is ring

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366036 A uniform disc of mass $m$ is fitted (pivoted smoothly) with a rod of mass $\frac{m}{2}$ . If the bottom of the rod is pulled with a velocity $v$ it moves without changing its angle of orientation and the disc rolls without sliding. The kinetic energy of the system(rod+disc) is
supporting img

1

\frac{m v^{2}}{4}

2

\frac{m v^{2}}{4}

3

\frac{15 m v^{2}}{16}

4

m v^{2}

Explanation:

Rolling can be considered as pure rotation about point of contact
$K_{r o l l i n g} = \frac{1}{2} I_{P} ω^{2}$
$= \frac{1}{2} (\frac{m R^{2}}{2} + m R^{2}) ω^{2}$
$= \frac{3}{4} m R^{2} ω^{2} (1)$
The rod translates with the velocity $v$ hence velocity of centre of disc will also be $v$
$v = ω R (2)$
From (1) and (2); $K_{rolling} = \frac{3}{4} m v^{2}$
Kinetic energy of the rod
$\begin{aligned} K_{rod} = \frac{1}{2} (\frac{m}{2}) v^{2} = \frac{m v^{2}}{4} \\ K_{total} = k_{rolling} + k_{rod} = m v^{2} \end{aligned}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366032 A loop and a disc roll without slipping with same linear velocity $v$ . The mass of the loop and the disc is same. If the total kinetic energy of the loop is 8 $J$ , find the kinetic energy of the disc

1

3 J

2

6 J

3

9 J

4

1 J

Explanation:

Total kinetic energy of loop
$= \frac{1}{2} I ω + \frac{1}{2} M v^{2}$ $= \frac{1}{2} (M R^{2} ω^{2}) + \frac{1}{2} M v^{2} = M v^{2} = 8 J (given)$
Total kinetic energy of disc
$= \frac{1}{2} I ω^{2} + \frac{1}{2} M v^{2}$
$= \frac{1}{2} (\frac{1}{2} M R^{2}) (\frac{v^{2}}{R^{2}}) + \frac{1}{2} M v^{2}$
$= \frac{3}{4} M v^{2} = \frac{3}{4} (8) = 6 J$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366033 A body is rolling down an inclined plane. If $K E$ of rotation is $40 %$ $K . E .$ of in translatory state, then the body is a

1 ring

2 cylinder

3 hollow ball

4 solid ball

Explanation:

$K E$ of rotation is $40 %$ of in translatory
$\begin{aligned} (K E)_{rot} & = 40 % \times (K E)_{trans} \\ \frac{1}{2} I ω^{2} & = \frac{40}{100} \times \frac{1}{2} m v^{2} \\ \Rightarrow I ω^{2} & = \frac{2}{5} m (ω r)^{2} [∵ v = ω r] \\ \Rightarrow I & = \frac{2}{5} m r^{2} \end{aligned}$
where, $I$ is moment of inertia.Thus, the body is a solid sphere (solid ball).

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366034 A ball rolls without slipping. The radius of gyration of the ball about an axis passing through its centre of mass is $K$ . If radius of the ball be $R$ , then the fraction of total enery associated with its rotational energy will be

1

\frac{K^{2}}{R^{2}}

2

\frac{K^{2}}{K^{2} + R^{2}}

3

\frac{R^{2}}{K^{2} + R^{2}}

4

\frac{K^{2} + R^{2}}{R^{2}}

Explanation:

Denote Kinetic energy $= E_{k}$
$\frac{{(E_{k})}_{r o t}}{E_{T}} = \frac{\frac{1}{2} m v^{2} (\frac{K^{2}}{R^{2}})}{\frac{1}{2} m v^{2} (1 + \frac{K^{2}}{R^{2}})} = \frac{K^{2}}{K^{2} + R^{2}}$
Where subscript (rot) is for rotational and subscript (T) for total. So, option (2) is correct.

MHTCET - 2021

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366035 A body is rolling without slipping on a horizontal surface and its rotational kinetic energy is equal to the translational kinetic energy. The body is

1 Sphere

2 Ring

3 Disc

4 Cylinder

Explanation:

$K_{R} = K_{T} \Rightarrow \frac{1}{2} m v^{2} (\frac{K^{2}}{R^{2}}) = \frac{1}{2} m v^{2} \Rightarrow \frac{K^{2}}{R^{2}} = 1$
i.e., the body is ring

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366036 A uniform disc of mass $m$ is fitted (pivoted smoothly) with a rod of mass $\frac{m}{2}$ . If the bottom of the rod is pulled with a velocity $v$ it moves without changing its angle of orientation and the disc rolls without sliding. The kinetic energy of the system(rod+disc) is
supporting img

1

\frac{m v^{2}}{4}

2

\frac{m v^{2}}{4}

3

\frac{15 m v^{2}}{16}

4

m v^{2}

Explanation:

Rolling can be considered as pure rotation about point of contact
$K_{r o l l i n g} = \frac{1}{2} I_{P} ω^{2}$
$= \frac{1}{2} (\frac{m R^{2}}{2} + m R^{2}) ω^{2}$
$= \frac{3}{4} m R^{2} ω^{2} (1)$
The rod translates with the velocity $v$ hence velocity of centre of disc will also be $v$
$v = ω R (2)$
From (1) and (2); $K_{rolling} = \frac{3}{4} m v^{2}$
Kinetic energy of the rod
$\begin{aligned} K_{rod} = \frac{1}{2} (\frac{m}{2}) v^{2} = \frac{m v^{2}}{4} \\ K_{total} = k_{rolling} + k_{rod} = m v^{2} \end{aligned}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366032 A loop and a disc roll without slipping with same linear velocity $v$ . The mass of the loop and the disc is same. If the total kinetic energy of the loop is 8 $J$ , find the kinetic energy of the disc

1

3 J

2

6 J

3

9 J

4

1 J

Explanation:

Total kinetic energy of loop
$= \frac{1}{2} I ω + \frac{1}{2} M v^{2}$ $= \frac{1}{2} (M R^{2} ω^{2}) + \frac{1}{2} M v^{2} = M v^{2} = 8 J (given)$
Total kinetic energy of disc
$= \frac{1}{2} I ω^{2} + \frac{1}{2} M v^{2}$
$= \frac{1}{2} (\frac{1}{2} M R^{2}) (\frac{v^{2}}{R^{2}}) + \frac{1}{2} M v^{2}$
$= \frac{3}{4} M v^{2} = \frac{3}{4} (8) = 6 J$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366033 A body is rolling down an inclined plane. If $K E$ of rotation is $40 %$ $K . E .$ of in translatory state, then the body is a

1 ring

2 cylinder

3 hollow ball

4 solid ball

Explanation:

$K E$ of rotation is $40 %$ of in translatory
$\begin{aligned} (K E)_{rot} & = 40 % \times (K E)_{trans} \\ \frac{1}{2} I ω^{2} & = \frac{40}{100} \times \frac{1}{2} m v^{2} \\ \Rightarrow I ω^{2} & = \frac{2}{5} m (ω r)^{2} [∵ v = ω r] \\ \Rightarrow I & = \frac{2}{5} m r^{2} \end{aligned}$
where, $I$ is moment of inertia.Thus, the body is a solid sphere (solid ball).

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366034 A ball rolls without slipping. The radius of gyration of the ball about an axis passing through its centre of mass is $K$ . If radius of the ball be $R$ , then the fraction of total enery associated with its rotational energy will be

1

\frac{K^{2}}{R^{2}}

2

\frac{K^{2}}{K^{2} + R^{2}}

3

\frac{R^{2}}{K^{2} + R^{2}}

4

\frac{K^{2} + R^{2}}{R^{2}}

Explanation:

Denote Kinetic energy $= E_{k}$
$\frac{{(E_{k})}_{r o t}}{E_{T}} = \frac{\frac{1}{2} m v^{2} (\frac{K^{2}}{R^{2}})}{\frac{1}{2} m v^{2} (1 + \frac{K^{2}}{R^{2}})} = \frac{K^{2}}{K^{2} + R^{2}}$
Where subscript (rot) is for rotational and subscript (T) for total. So, option (2) is correct.

MHTCET - 2021

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366035 A body is rolling without slipping on a horizontal surface and its rotational kinetic energy is equal to the translational kinetic energy. The body is

1 Sphere

2 Ring

3 Disc

4 Cylinder

Explanation:

$K_{R} = K_{T} \Rightarrow \frac{1}{2} m v^{2} (\frac{K^{2}}{R^{2}}) = \frac{1}{2} m v^{2} \Rightarrow \frac{K^{2}}{R^{2}} = 1$
i.e., the body is ring

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366036 A uniform disc of mass $m$ is fitted (pivoted smoothly) with a rod of mass $\frac{m}{2}$ . If the bottom of the rod is pulled with a velocity $v$ it moves without changing its angle of orientation and the disc rolls without sliding. The kinetic energy of the system(rod+disc) is
supporting img

1

\frac{m v^{2}}{4}

2

\frac{m v^{2}}{4}

3

\frac{15 m v^{2}}{16}

4

m v^{2}

Explanation:

Rolling can be considered as pure rotation about point of contact
$K_{r o l l i n g} = \frac{1}{2} I_{P} ω^{2}$
$= \frac{1}{2} (\frac{m R^{2}}{2} + m R^{2}) ω^{2}$
$= \frac{3}{4} m R^{2} ω^{2} (1)$
The rod translates with the velocity $v$ hence velocity of centre of disc will also be $v$
$v = ω R (2)$
From (1) and (2); $K_{rolling} = \frac{3}{4} m v^{2}$
Kinetic energy of the rod
$\begin{aligned} K_{rod} = \frac{1}{2} (\frac{m}{2}) v^{2} = \frac{m v^{2}}{4} \\ K_{total} = k_{rolling} + k_{rod} = m v^{2} \end{aligned}$

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