QBManager

Rotational Motion

150310 A solid metal ring and a disc of same radius and mass are rotating about their diameters with same angular frequency. The ratio of their respective rotational kinetic energy values is

1

1 : 1

2

1 : 2

3

2 : 1

4

1 : 4

5

4 : 1

Explanation:

C Let radius of ring $=$ radius of disc $= r$
Mass of ring $=$ mass of disc $= m$
Angular frequency of ring $=$ angular frequency of disc $=$ $ω$
Kinetic energy of solid metal ring
$K_{ring} = \frac{1}{2} I_{ring} ω_{ring}^{2}$
$K_{ring} = \frac{1}{2} \times {mr}^{2} ω^{2}$
Kinetic energy of disc
$K_{disc} = \frac{1}{2} \times I_{disc} ω_{disc}^{2}$
$K_{disc} = \frac{1}{2} \times (\frac{{mr}^{2}}{2}) ω^{2}$
$K_{disc} = \frac{1}{4} \times {mr}^{2} ω^{2}$
The ratio of their respective rotational kinetic energy is
$\frac{K_{ring}}{K_{disc}} = \frac{\frac{1}{2} {mr}^{2} ω^{2}}{\frac{1}{4} {mr}^{2} ω^{2}}$
$\frac{K_{ring}}{K_{disc}} = \frac{2}{1}$

Kerala CEE 2021

Rotational Motion

150313 As solid of mass ' $M$ ' and radius ' $R$ ' spins about an axis passing through its centre making 600 rpm. Its kinetic energy of rotation is

1

\frac{2 π^{2}}{5} MR

2

\frac{2 π}{5} M^{2} R^{2}

3

80 π MR

4

80 π^{2} {MR}^{2}

Explanation:

D Given,
Angular speed $ω = 2 π N = 2 π \times \frac{600}{60} = 20 π rad / s$
$∴$ Rotational $K \cdot E$ . of the sphere $= \frac{1}{2} I ω^{2}$
$= \frac{1}{2} \cdot \frac{2}{5} {MR}^{2} \cdot (20 π)^{2} (∵ I = \frac{2}{5} {MR}^{2})$
$= \frac{1}{5} {MR}^{2} \times 400 π^{2}$
$K \cdot E = 80 π^{2} {MR}^{2}$

AP EAMCET-20.08.2021

Rotational Motion

150314 A wheel of mass $10 kg$ , radius of gyration 50 $cm$ is rotating at $300 rpm$ . The rotational kinetic energy of the wheel is

1

625 J

2

1000 J

3

1250 J

4

1500 J

Explanation:

C Given,
Mass of wheel $m) = 10 kg$
Radius of gyration $k) = 50 cm = 0.5 m$
No of Rotation of wheel $N) = 300 rpm$
We know that,
$ω = \frac{2 π N}{60} = \frac{2 \times π \times 300}{60} = 10 π rad / s$
K.E. $= \frac{1}{2} I ω^{2} = \frac{1}{2} {mk}^{2} ω^{2} = \frac{1}{2} \times 10 \times (0.5)^{2} \times (10 π)^{2}$
$K.E. = 1232.45 \approx 1250 J$

AP EAMCET-06.09.2021

Rotational Motion

150315 A thin uniform rod of mass $1 kg$ and length $1 m$ is hanged at one end to the ground floor. It originally stands vertically and allowed to fall ground. If the rod hits the ground with angular speed $ω$ then the correct statement is (Assume $g = 10 m / s^{2}$

1

ω = \sqrt{30} rad / s

2

ω = \sqrt{20} rad / s

3

ω = 5 rad / s

4

ω = 6 rad / s

Explanation:

A Given, mass of $rod (M) = 1 kg$ , length of $rod$ $L) = 1 m$ , angular speed $= ω, g = 10 m / s^{2}$
According to law of conservation of energy-
$Mg (\frac{L}{2}) + 0 = 0 + \frac{1}{2} I ω^{2}$
$Mg \frac{L}{2} = \frac{1}{2} (\frac{{ML}^{2}}{3}) ω^{2} [I = \frac{{ML}^{2}}{3}]$
$\frac{3 g}{L} = ω^{2}$
$ω = \sqrt{\frac{3 g}{L}}$
Putting the value of ' $L$ ' and ' $g$ ', we get-
$ω = \sqrt{\frac{3 \times 10}{1}} = \sqrt{30}$
$ω = \sqrt{30} rad / \sec$

TS EAMCET 04.08.2021

Rotational Motion

150316 A molecule consists of two atoms each of mass ' $m$ ' and separated by distance ' $d$ '. At room temperature the average rotational kinetic energy is ' $E$ ', then its angular frequency is

1

\frac{2}{d} \sqrt{\frac{E}{m}}

2

\sqrt{\frac{m}{E d}}

3

\frac{d}{2} \sqrt{\frac{m}{E}}

4

\sqrt{\frac{Ed}{m}}

Explanation:

A Given, mass of both atoms $= m$ separation distance $= d$ rotational kinetic energy $= E$ Angular frequency $=$ ?
The moment of inertia of the molecule,
$I = m {(\frac{d}{2})}^{2} + m {(\frac{d}{2})}^{2}$
$I = \frac{{md}^{2}}{2}$
Rotational Kinetic energy $(E) = \frac{1}{2} I ω^{2}$
Putting the value of I from equation (i), we get -
$E = \frac{1}{2} \times \frac{{md}^{2}}{2} \times ω^{2}$
$ω^{2} = \frac{4 E}{{md}^{2}}$
$ω = \sqrt{\frac{4 E}{{md}^{2}}}$
Angular frequency $ω) = \frac{2}{d} \sqrt{\frac{E}{m}}$

MHT-CET 2020

Rotational Motion

150310 A solid metal ring and a disc of same radius and mass are rotating about their diameters with same angular frequency. The ratio of their respective rotational kinetic energy values is

1

1 : 1

2

1 : 2

3

2 : 1

4

1 : 4

5

4 : 1

Explanation:

C Let radius of ring $=$ radius of disc $= r$
Mass of ring $=$ mass of disc $= m$
Angular frequency of ring $=$ angular frequency of disc $=$ $ω$
Kinetic energy of solid metal ring
$K_{ring} = \frac{1}{2} I_{ring} ω_{ring}^{2}$
$K_{ring} = \frac{1}{2} \times {mr}^{2} ω^{2}$
Kinetic energy of disc
$K_{disc} = \frac{1}{2} \times I_{disc} ω_{disc}^{2}$
$K_{disc} = \frac{1}{2} \times (\frac{{mr}^{2}}{2}) ω^{2}$
$K_{disc} = \frac{1}{4} \times {mr}^{2} ω^{2}$
The ratio of their respective rotational kinetic energy is
$\frac{K_{ring}}{K_{disc}} = \frac{\frac{1}{2} {mr}^{2} ω^{2}}{\frac{1}{4} {mr}^{2} ω^{2}}$
$\frac{K_{ring}}{K_{disc}} = \frac{2}{1}$

Kerala CEE 2021

Rotational Motion

150313 As solid of mass ' $M$ ' and radius ' $R$ ' spins about an axis passing through its centre making 600 rpm. Its kinetic energy of rotation is

1

\frac{2 π^{2}}{5} MR

2

\frac{2 π}{5} M^{2} R^{2}

3

80 π MR

4

80 π^{2} {MR}^{2}

Explanation:

D Given,
Angular speed $ω = 2 π N = 2 π \times \frac{600}{60} = 20 π rad / s$
$∴$ Rotational $K \cdot E$ . of the sphere $= \frac{1}{2} I ω^{2}$
$= \frac{1}{2} \cdot \frac{2}{5} {MR}^{2} \cdot (20 π)^{2} (∵ I = \frac{2}{5} {MR}^{2})$
$= \frac{1}{5} {MR}^{2} \times 400 π^{2}$
$K \cdot E = 80 π^{2} {MR}^{2}$

AP EAMCET-20.08.2021

Rotational Motion

150314 A wheel of mass $10 kg$ , radius of gyration 50 $cm$ is rotating at $300 rpm$ . The rotational kinetic energy of the wheel is

1

625 J

2

1000 J

3

1250 J

4

1500 J

Explanation:

C Given,
Mass of wheel $m) = 10 kg$
Radius of gyration $k) = 50 cm = 0.5 m$
No of Rotation of wheel $N) = 300 rpm$
We know that,
$ω = \frac{2 π N}{60} = \frac{2 \times π \times 300}{60} = 10 π rad / s$
K.E. $= \frac{1}{2} I ω^{2} = \frac{1}{2} {mk}^{2} ω^{2} = \frac{1}{2} \times 10 \times (0.5)^{2} \times (10 π)^{2}$
$K.E. = 1232.45 \approx 1250 J$

AP EAMCET-06.09.2021

Rotational Motion

150315 A thin uniform rod of mass $1 kg$ and length $1 m$ is hanged at one end to the ground floor. It originally stands vertically and allowed to fall ground. If the rod hits the ground with angular speed $ω$ then the correct statement is (Assume $g = 10 m / s^{2}$

1

ω = \sqrt{30} rad / s

2

ω = \sqrt{20} rad / s

3

ω = 5 rad / s

4

ω = 6 rad / s

Explanation:

A Given, mass of $rod (M) = 1 kg$ , length of $rod$ $L) = 1 m$ , angular speed $= ω, g = 10 m / s^{2}$
According to law of conservation of energy-
$Mg (\frac{L}{2}) + 0 = 0 + \frac{1}{2} I ω^{2}$
$Mg \frac{L}{2} = \frac{1}{2} (\frac{{ML}^{2}}{3}) ω^{2} [I = \frac{{ML}^{2}}{3}]$
$\frac{3 g}{L} = ω^{2}$
$ω = \sqrt{\frac{3 g}{L}}$
Putting the value of ' $L$ ' and ' $g$ ', we get-
$ω = \sqrt{\frac{3 \times 10}{1}} = \sqrt{30}$
$ω = \sqrt{30} rad / \sec$

TS EAMCET 04.08.2021

Rotational Motion

150316 A molecule consists of two atoms each of mass ' $m$ ' and separated by distance ' $d$ '. At room temperature the average rotational kinetic energy is ' $E$ ', then its angular frequency is

1

\frac{2}{d} \sqrt{\frac{E}{m}}

2

\sqrt{\frac{m}{E d}}

3

\frac{d}{2} \sqrt{\frac{m}{E}}

4

\sqrt{\frac{Ed}{m}}

Explanation:

A Given, mass of both atoms $= m$ separation distance $= d$ rotational kinetic energy $= E$ Angular frequency $=$ ?
The moment of inertia of the molecule,
$I = m {(\frac{d}{2})}^{2} + m {(\frac{d}{2})}^{2}$
$I = \frac{{md}^{2}}{2}$
Rotational Kinetic energy $(E) = \frac{1}{2} I ω^{2}$
Putting the value of I from equation (i), we get -
$E = \frac{1}{2} \times \frac{{md}^{2}}{2} \times ω^{2}$
$ω^{2} = \frac{4 E}{{md}^{2}}$
$ω = \sqrt{\frac{4 E}{{md}^{2}}}$
Angular frequency $ω) = \frac{2}{d} \sqrt{\frac{E}{m}}$

MHT-CET 2020

Rotational Motion

150310 A solid metal ring and a disc of same radius and mass are rotating about their diameters with same angular frequency. The ratio of their respective rotational kinetic energy values is

1

1 : 1

2

1 : 2

3

2 : 1

4

1 : 4

5

4 : 1

Explanation:

C Let radius of ring $=$ radius of disc $= r$
Mass of ring $=$ mass of disc $= m$
Angular frequency of ring $=$ angular frequency of disc $=$ $ω$
Kinetic energy of solid metal ring
$K_{ring} = \frac{1}{2} I_{ring} ω_{ring}^{2}$
$K_{ring} = \frac{1}{2} \times {mr}^{2} ω^{2}$
Kinetic energy of disc
$K_{disc} = \frac{1}{2} \times I_{disc} ω_{disc}^{2}$
$K_{disc} = \frac{1}{2} \times (\frac{{mr}^{2}}{2}) ω^{2}$
$K_{disc} = \frac{1}{4} \times {mr}^{2} ω^{2}$
The ratio of their respective rotational kinetic energy is
$\frac{K_{ring}}{K_{disc}} = \frac{\frac{1}{2} {mr}^{2} ω^{2}}{\frac{1}{4} {mr}^{2} ω^{2}}$
$\frac{K_{ring}}{K_{disc}} = \frac{2}{1}$

Kerala CEE 2021

Rotational Motion

150313 As solid of mass ' $M$ ' and radius ' $R$ ' spins about an axis passing through its centre making 600 rpm. Its kinetic energy of rotation is

1

\frac{2 π^{2}}{5} MR

2

\frac{2 π}{5} M^{2} R^{2}

3

80 π MR

4

80 π^{2} {MR}^{2}

Explanation:

D Given,
Angular speed $ω = 2 π N = 2 π \times \frac{600}{60} = 20 π rad / s$
$∴$ Rotational $K \cdot E$ . of the sphere $= \frac{1}{2} I ω^{2}$
$= \frac{1}{2} \cdot \frac{2}{5} {MR}^{2} \cdot (20 π)^{2} (∵ I = \frac{2}{5} {MR}^{2})$
$= \frac{1}{5} {MR}^{2} \times 400 π^{2}$
$K \cdot E = 80 π^{2} {MR}^{2}$

AP EAMCET-20.08.2021

Rotational Motion

150314 A wheel of mass $10 kg$ , radius of gyration 50 $cm$ is rotating at $300 rpm$ . The rotational kinetic energy of the wheel is

1

625 J

2

1000 J

3

1250 J

4

1500 J

Explanation:

C Given,
Mass of wheel $m) = 10 kg$
Radius of gyration $k) = 50 cm = 0.5 m$
No of Rotation of wheel $N) = 300 rpm$
We know that,
$ω = \frac{2 π N}{60} = \frac{2 \times π \times 300}{60} = 10 π rad / s$
K.E. $= \frac{1}{2} I ω^{2} = \frac{1}{2} {mk}^{2} ω^{2} = \frac{1}{2} \times 10 \times (0.5)^{2} \times (10 π)^{2}$
$K.E. = 1232.45 \approx 1250 J$

AP EAMCET-06.09.2021

Rotational Motion

150315 A thin uniform rod of mass $1 kg$ and length $1 m$ is hanged at one end to the ground floor. It originally stands vertically and allowed to fall ground. If the rod hits the ground with angular speed $ω$ then the correct statement is (Assume $g = 10 m / s^{2}$

1

ω = \sqrt{30} rad / s

2

ω = \sqrt{20} rad / s

3

ω = 5 rad / s

4

ω = 6 rad / s

Explanation:

A Given, mass of $rod (M) = 1 kg$ , length of $rod$ $L) = 1 m$ , angular speed $= ω, g = 10 m / s^{2}$
According to law of conservation of energy-
$Mg (\frac{L}{2}) + 0 = 0 + \frac{1}{2} I ω^{2}$
$Mg \frac{L}{2} = \frac{1}{2} (\frac{{ML}^{2}}{3}) ω^{2} [I = \frac{{ML}^{2}}{3}]$
$\frac{3 g}{L} = ω^{2}$
$ω = \sqrt{\frac{3 g}{L}}$
Putting the value of ' $L$ ' and ' $g$ ', we get-
$ω = \sqrt{\frac{3 \times 10}{1}} = \sqrt{30}$
$ω = \sqrt{30} rad / \sec$

TS EAMCET 04.08.2021

Rotational Motion

150316 A molecule consists of two atoms each of mass ' $m$ ' and separated by distance ' $d$ '. At room temperature the average rotational kinetic energy is ' $E$ ', then its angular frequency is

1

\frac{2}{d} \sqrt{\frac{E}{m}}

2

\sqrt{\frac{m}{E d}}

3

\frac{d}{2} \sqrt{\frac{m}{E}}

4

\sqrt{\frac{Ed}{m}}

Explanation:

A Given, mass of both atoms $= m$ separation distance $= d$ rotational kinetic energy $= E$ Angular frequency $=$ ?
The moment of inertia of the molecule,
$I = m {(\frac{d}{2})}^{2} + m {(\frac{d}{2})}^{2}$
$I = \frac{{md}^{2}}{2}$
Rotational Kinetic energy $(E) = \frac{1}{2} I ω^{2}$
Putting the value of I from equation (i), we get -
$E = \frac{1}{2} \times \frac{{md}^{2}}{2} \times ω^{2}$
$ω^{2} = \frac{4 E}{{md}^{2}}$
$ω = \sqrt{\frac{4 E}{{md}^{2}}}$
Angular frequency $ω) = \frac{2}{d} \sqrt{\frac{E}{m}}$

MHT-CET 2020

Rotational Motion

150310 A solid metal ring and a disc of same radius and mass are rotating about their diameters with same angular frequency. The ratio of their respective rotational kinetic energy values is

1

1 : 1

2

1 : 2

3

2 : 1

4

1 : 4

5

4 : 1

Explanation:

C Let radius of ring $=$ radius of disc $= r$
Mass of ring $=$ mass of disc $= m$
Angular frequency of ring $=$ angular frequency of disc $=$ $ω$
Kinetic energy of solid metal ring
$K_{ring} = \frac{1}{2} I_{ring} ω_{ring}^{2}$
$K_{ring} = \frac{1}{2} \times {mr}^{2} ω^{2}$
Kinetic energy of disc
$K_{disc} = \frac{1}{2} \times I_{disc} ω_{disc}^{2}$
$K_{disc} = \frac{1}{2} \times (\frac{{mr}^{2}}{2}) ω^{2}$
$K_{disc} = \frac{1}{4} \times {mr}^{2} ω^{2}$
The ratio of their respective rotational kinetic energy is
$\frac{K_{ring}}{K_{disc}} = \frac{\frac{1}{2} {mr}^{2} ω^{2}}{\frac{1}{4} {mr}^{2} ω^{2}}$
$\frac{K_{ring}}{K_{disc}} = \frac{2}{1}$

Kerala CEE 2021

Rotational Motion

150313 As solid of mass ' $M$ ' and radius ' $R$ ' spins about an axis passing through its centre making 600 rpm. Its kinetic energy of rotation is

1

\frac{2 π^{2}}{5} MR

2

\frac{2 π}{5} M^{2} R^{2}

3

80 π MR

4

80 π^{2} {MR}^{2}

Explanation:

D Given,
Angular speed $ω = 2 π N = 2 π \times \frac{600}{60} = 20 π rad / s$
$∴$ Rotational $K \cdot E$ . of the sphere $= \frac{1}{2} I ω^{2}$
$= \frac{1}{2} \cdot \frac{2}{5} {MR}^{2} \cdot (20 π)^{2} (∵ I = \frac{2}{5} {MR}^{2})$
$= \frac{1}{5} {MR}^{2} \times 400 π^{2}$
$K \cdot E = 80 π^{2} {MR}^{2}$

AP EAMCET-20.08.2021

Rotational Motion

150314 A wheel of mass $10 kg$ , radius of gyration 50 $cm$ is rotating at $300 rpm$ . The rotational kinetic energy of the wheel is

1

625 J

2

1000 J

3

1250 J

4

1500 J

Explanation:

C Given,
Mass of wheel $m) = 10 kg$
Radius of gyration $k) = 50 cm = 0.5 m$
No of Rotation of wheel $N) = 300 rpm$
We know that,
$ω = \frac{2 π N}{60} = \frac{2 \times π \times 300}{60} = 10 π rad / s$
K.E. $= \frac{1}{2} I ω^{2} = \frac{1}{2} {mk}^{2} ω^{2} = \frac{1}{2} \times 10 \times (0.5)^{2} \times (10 π)^{2}$
$K.E. = 1232.45 \approx 1250 J$

AP EAMCET-06.09.2021

Rotational Motion

150315 A thin uniform rod of mass $1 kg$ and length $1 m$ is hanged at one end to the ground floor. It originally stands vertically and allowed to fall ground. If the rod hits the ground with angular speed $ω$ then the correct statement is (Assume $g = 10 m / s^{2}$

1

ω = \sqrt{30} rad / s

2

ω = \sqrt{20} rad / s

3

ω = 5 rad / s

4

ω = 6 rad / s

Explanation:

A Given, mass of $rod (M) = 1 kg$ , length of $rod$ $L) = 1 m$ , angular speed $= ω, g = 10 m / s^{2}$
According to law of conservation of energy-
$Mg (\frac{L}{2}) + 0 = 0 + \frac{1}{2} I ω^{2}$
$Mg \frac{L}{2} = \frac{1}{2} (\frac{{ML}^{2}}{3}) ω^{2} [I = \frac{{ML}^{2}}{3}]$
$\frac{3 g}{L} = ω^{2}$
$ω = \sqrt{\frac{3 g}{L}}$
Putting the value of ' $L$ ' and ' $g$ ', we get-
$ω = \sqrt{\frac{3 \times 10}{1}} = \sqrt{30}$
$ω = \sqrt{30} rad / \sec$

TS EAMCET 04.08.2021

Rotational Motion

150316 A molecule consists of two atoms each of mass ' $m$ ' and separated by distance ' $d$ '. At room temperature the average rotational kinetic energy is ' $E$ ', then its angular frequency is

1

\frac{2}{d} \sqrt{\frac{E}{m}}

2

\sqrt{\frac{m}{E d}}

3

\frac{d}{2} \sqrt{\frac{m}{E}}

4

\sqrt{\frac{Ed}{m}}

Explanation:

A Given, mass of both atoms $= m$ separation distance $= d$ rotational kinetic energy $= E$ Angular frequency $=$ ?
The moment of inertia of the molecule,
$I = m {(\frac{d}{2})}^{2} + m {(\frac{d}{2})}^{2}$
$I = \frac{{md}^{2}}{2}$
Rotational Kinetic energy $(E) = \frac{1}{2} I ω^{2}$
Putting the value of I from equation (i), we get -
$E = \frac{1}{2} \times \frac{{md}^{2}}{2} \times ω^{2}$
$ω^{2} = \frac{4 E}{{md}^{2}}$
$ω = \sqrt{\frac{4 E}{{md}^{2}}}$
Angular frequency $ω) = \frac{2}{d} \sqrt{\frac{E}{m}}$

MHT-CET 2020

Rotational Motion

150310 A solid metal ring and a disc of same radius and mass are rotating about their diameters with same angular frequency. The ratio of their respective rotational kinetic energy values is

1

1 : 1

2

1 : 2

3

2 : 1

4

1 : 4

5

4 : 1

Explanation:

C Let radius of ring $=$ radius of disc $= r$
Mass of ring $=$ mass of disc $= m$
Angular frequency of ring $=$ angular frequency of disc $=$ $ω$
Kinetic energy of solid metal ring
$K_{ring} = \frac{1}{2} I_{ring} ω_{ring}^{2}$
$K_{ring} = \frac{1}{2} \times {mr}^{2} ω^{2}$
Kinetic energy of disc
$K_{disc} = \frac{1}{2} \times I_{disc} ω_{disc}^{2}$
$K_{disc} = \frac{1}{2} \times (\frac{{mr}^{2}}{2}) ω^{2}$
$K_{disc} = \frac{1}{4} \times {mr}^{2} ω^{2}$
The ratio of their respective rotational kinetic energy is
$\frac{K_{ring}}{K_{disc}} = \frac{\frac{1}{2} {mr}^{2} ω^{2}}{\frac{1}{4} {mr}^{2} ω^{2}}$
$\frac{K_{ring}}{K_{disc}} = \frac{2}{1}$

Kerala CEE 2021

Rotational Motion

150313 As solid of mass ' $M$ ' and radius ' $R$ ' spins about an axis passing through its centre making 600 rpm. Its kinetic energy of rotation is

1

\frac{2 π^{2}}{5} MR

2

\frac{2 π}{5} M^{2} R^{2}

3

80 π MR

4

80 π^{2} {MR}^{2}

Explanation:

D Given,
Angular speed $ω = 2 π N = 2 π \times \frac{600}{60} = 20 π rad / s$
$∴$ Rotational $K \cdot E$ . of the sphere $= \frac{1}{2} I ω^{2}$
$= \frac{1}{2} \cdot \frac{2}{5} {MR}^{2} \cdot (20 π)^{2} (∵ I = \frac{2}{5} {MR}^{2})$
$= \frac{1}{5} {MR}^{2} \times 400 π^{2}$
$K \cdot E = 80 π^{2} {MR}^{2}$

AP EAMCET-20.08.2021

Rotational Motion

150314 A wheel of mass $10 kg$ , radius of gyration 50 $cm$ is rotating at $300 rpm$ . The rotational kinetic energy of the wheel is

1

625 J

2

1000 J

3

1250 J

4

1500 J

Explanation:

C Given,
Mass of wheel $m) = 10 kg$
Radius of gyration $k) = 50 cm = 0.5 m$
No of Rotation of wheel $N) = 300 rpm$
We know that,
$ω = \frac{2 π N}{60} = \frac{2 \times π \times 300}{60} = 10 π rad / s$
K.E. $= \frac{1}{2} I ω^{2} = \frac{1}{2} {mk}^{2} ω^{2} = \frac{1}{2} \times 10 \times (0.5)^{2} \times (10 π)^{2}$
$K.E. = 1232.45 \approx 1250 J$

AP EAMCET-06.09.2021

Rotational Motion

150315 A thin uniform rod of mass $1 kg$ and length $1 m$ is hanged at one end to the ground floor. It originally stands vertically and allowed to fall ground. If the rod hits the ground with angular speed $ω$ then the correct statement is (Assume $g = 10 m / s^{2}$

1

ω = \sqrt{30} rad / s

2

ω = \sqrt{20} rad / s

3

ω = 5 rad / s

4

ω = 6 rad / s

Explanation:

A Given, mass of $rod (M) = 1 kg$ , length of $rod$ $L) = 1 m$ , angular speed $= ω, g = 10 m / s^{2}$
According to law of conservation of energy-
$Mg (\frac{L}{2}) + 0 = 0 + \frac{1}{2} I ω^{2}$
$Mg \frac{L}{2} = \frac{1}{2} (\frac{{ML}^{2}}{3}) ω^{2} [I = \frac{{ML}^{2}}{3}]$
$\frac{3 g}{L} = ω^{2}$
$ω = \sqrt{\frac{3 g}{L}}$
Putting the value of ' $L$ ' and ' $g$ ', we get-
$ω = \sqrt{\frac{3 \times 10}{1}} = \sqrt{30}$
$ω = \sqrt{30} rad / \sec$

TS EAMCET 04.08.2021

Rotational Motion

150316 A molecule consists of two atoms each of mass ' $m$ ' and separated by distance ' $d$ '. At room temperature the average rotational kinetic energy is ' $E$ ', then its angular frequency is

1

\frac{2}{d} \sqrt{\frac{E}{m}}

2

\sqrt{\frac{m}{E d}}

3

\frac{d}{2} \sqrt{\frac{m}{E}}

4

\sqrt{\frac{Ed}{m}}

Explanation:

A Given, mass of both atoms $= m$ separation distance $= d$ rotational kinetic energy $= E$ Angular frequency $=$ ?
The moment of inertia of the molecule,
$I = m {(\frac{d}{2})}^{2} + m {(\frac{d}{2})}^{2}$
$I = \frac{{md}^{2}}{2}$
Rotational Kinetic energy $(E) = \frac{1}{2} I ω^{2}$
Putting the value of I from equation (i), we get -
$E = \frac{1}{2} \times \frac{{md}^{2}}{2} \times ω^{2}$
$ω^{2} = \frac{4 E}{{md}^{2}}$
$ω = \sqrt{\frac{4 E}{{md}^{2}}}$
Angular frequency $ω) = \frac{2}{d} \sqrt{\frac{E}{m}}$

MHT-CET 2020

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation: