63 + Interview Questions in ROTATIONAL DYNAMICS IN GENERAL AWARENESS Page 2 InterviewSolution

51.	If L is the angular momentum and I is the moment of inertia of a rotating body, then \(\frac{L^2}{2I}\) represents its (A) rotational PE (B) total energy (C) rotational KE (D) translational KE.
Answer» (C) rotational KE

Discussion

52.	A track for a certain motor sport event is in the form of a circle and banked at an angle θ. For a car driven in a circle of radius r along the track at the optimum speed, the periodic time is(A) \(\sqrt{\frac{r}g}\)(B) 2\(\pi\)\(\sqrt{\frac{r}g}\)(C) 2\(\pi\)\(\sqrt{\frac{r}{g\,tan\theta}}\) (D) 2\(\pi\)\(\sqrt{\frac{r\,tan\theta}{g}}\)
Answer» Correct Option is (C) \(2 \pi \sqrt{\frac{r}{g \ tan \theta}}\) We know that T = \(T =\frac{2 \pi r}{v}\) \(\because v = \sqrt{rg \ tan \theta} \) \(T = \frac{2 \pi r}{\sqrt{rg \ tan \theta}}\) \( T = 2 \pi \sqrt{\frac{r}{g \ tan \theta}}\) (C) 2\(\pi\)\(\sqrt{\frac{r}{g\,tan\theta}}\)

52.

A track for a certain motor sport event is in the form of a circle and banked at an angle θ. For a car driven in a circle of radius r along the track at the optimum speed, the periodic time is(A) \(\sqrt{\frac{r}g}\)(B) 2\(\pi\)\(\sqrt{\frac{r}g}\)(C) 2\(\pi\)\(\sqrt{\frac{r}{g\,tan\theta}}\) (D) 2\(\pi\)\(\sqrt{\frac{r\,tan\theta}{g}}\)

Answer»

Correct Option is (C) \(2 \pi \sqrt{\frac{r}{g \ tan \theta}}\)

We know that T = \(T =\frac{2 \pi r}{v}\)

\(\because v = \sqrt{rg \ tan \theta} \)

\(T = \frac{2 \pi r}{\sqrt{rg \ tan \theta}}\)

\( T = 2 \pi \sqrt{\frac{r}{g \ tan \theta}}\)

(C) 2\(\pi\)\(\sqrt{\frac{r}{g\,tan\theta}}\)

Discussion

53.	A body, tied to a string, performs circular motion in a vertical plane such that the tension in the string is zero at the highest point. What is the angular speed of the body at the 1. highest position 2. lowest position ?
Answer» 1.\(\sqrt{g/r}\) 2. \(\sqrt{5g/r}\) in the usual notation.

53.

A body, tied to a string, performs circular motion in a vertical plane such that the tension in the string is zero at the highest point. What is the angular speed of the body at the 1. highest position 2. lowest position ?

Answer»

1.\(\sqrt{g/r}\)

2. \(\sqrt{5g/r}\) in the usual notation.

Discussion

54.	A body, tied to a string, performs circular motion in a vertical plane such that the tension in the string is zero at the highest point. What is the linear speed of the body at the1. lowest position 2. highest position ?
Answer» 1.\(\sqrt{5 r g}\) 2. \(\sqrt{r g}\) in the usual notation.

54.

A body, tied to a string, performs circular motion in a vertical plane such that the tension in the string is zero at the highest point. What is the linear speed of the body at the1. lowest position 2. highest position ?

Answer»

1.\(\sqrt{5 r g}\)

2. \(\sqrt{r g}\) in the usual notation.

Discussion

55.	A bucket of water is whirled in a vertical circle at an arm’s length. Find the minimum speed at the top so that no water spills out. Also find the corresponding angular speed. [Assume r = 0.75 m]
Answer» Data : r = 0.75 m, g = 10 m/s² At the highest point the minimum speed required is v = \(\sqrt{rg}\) = \(\sqrt{0.75\times 10}\) = 2.738 ω = \(\frac{v}r\)= \(\frac{2.738}{0.75}\) = 3.651 rad/s

55.

A bucket of water is whirled in a vertical circle at an arm’s length. Find the minimum speed at the top so that no water spills out. Also find the corresponding angular speed. [Assume r = 0.75 m]

Answer»

Data : r = 0.75 m, g = 10 m/s²

At the highest point the minimum speed

required is v = \(\sqrt{rg}\) = \(\sqrt{0.75\times 10}\) = 2.738

ω = \(\frac{v}r\)= \(\frac{2.738}{0.75}\) = 3.651 rad/s

Discussion

56.	Attach a body of suitable mass to a spring balance so that it stretches by about half its capacity. Now whirl the spring balance so that the body performs a horizontal circular motion. You will notice that the balance now reads more for the same body. Can you explain this ?
Answer» Due to outward centrifugal force.

Discussion

57.	A disc of mass 4 kg rolls on a horizontal surface. If its linear speed is 3 m/s, what is its total kinetic energy?
Answer» Total kinetic energy of the disc = 3/4 Mv² = 3/4 x 4 x (3)² = 27 j

57.

A disc of mass 4 kg rolls on a horizontal surface. If its linear speed is 3 m/s, what is its total kinetic energy?

Answer»

Total kinetic energy of the disc

= 3/4 Mv² = 3/4 x 4 x (3)² = 27 j

Discussion

58.	A uniform solid sphere of mass 10 kg rolls on a horizontal surface. If its linear speed is 2 m/s, what is its total kinetic energy?
Answer» Total kinetic energy of the sphere = 7/10 Mv²= 7/10 x 10 x (2)² = 28 j

58.

A uniform solid sphere of mass 10 kg rolls on a horizontal surface. If its linear speed is 2 m/s, what is its total kinetic energy?

Answer»

Total kinetic energy of the sphere

= 7/10 Mv²= 7/10 x 10 x (2)² = 28 j

Discussion

59.	A spherical shell rolls down a plane inclined at 30° to the horizontal. What is its acceleration ?
Answer» The acceleration of the spherical shell, a = 3/5g sin \(\theta\) = 0.5g sin 30° = 0.6g × 1/2 = 0.3g

Discussion

60.	Certain stars are believed to be rotating at about 1 rot/s. If such a star has a diameter of 40 km, what is the linear speed of a point on the equator of the star?
Answer» Data : d = 40 km, f = 1 rot/s ∴ r = \(\frac{d}2\) = \(\frac{40\,km}2\) = 20 km = 2 x 10⁴ m Linear speed, v = ωr = (2πf)r = (2 × 3.142 × 1)(2 × 10⁴) = 6.284 × 2 × 10⁴ = 1.257 × 10 m/s (or 125.6 km/s)

60.

Certain stars are believed to be rotating at about 1 rot/s. If such a star has a diameter of 40 km, what is the linear speed of a point on the equator of the star?

Answer»

Data : d = 40 km, f = 1 rot/s

∴ r = \(\frac{d}2\) = \(\frac{40\,km}2\) = 20 km = 2 x 10⁴ m

Linear speed, v = ωr = (2πf)r

= (2 × 3.142 × 1)(2 × 10⁴)

= 6.284 × 2 × 10⁴

= 1.257 × 10 m/s (or 125.6 km/s)

Discussion

61.	An object of mass 0.5 kg is tied to a string and revolved in a horizontal circle of radius 1 m. If the breaking tension of the string is 50 N, what is the maximum speed the object can have?
Answer» Data : m = 0.5 kg, r = 1 m, F = 50 N The maximum centripetal force that can be applied is equal to the breaking tension. \(\therefore\) \(\frac{mv^2}r\) = F \(\therefore\) v = \(\sqrt{\frac{Fr}m}\) = \(\sqrt{\frac{50 \times1}{0.5}}\) = 10 m/s This is the maximum speed the object can have.

61.

An object of mass 0.5 kg is tied to a string and revolved in a horizontal circle of radius 1 m. If the breaking tension of the string is 50 N, what is the maximum speed the object can have?

Answer»

Data : m = 0.5 kg, r = 1 m, F = 50 N The maximum centripetal force that can be applied is equal to the breaking tension.

\(\therefore\) \(\frac{mv^2}r\) = F

\(\therefore\) v = \(\sqrt{\frac{Fr}m}\) = \(\sqrt{\frac{50 \times1}{0.5}}\) = 10 m/s

This is the maximum speed the object can have.

Discussion

62.	A body of mass 100 grams is tied to one end of a string and revolved along a circular path in the horizontal plane. The radius of the circle is 50 cm. If the body revolves with a constant angular speed of 20 rad/s, find the1. period of revolution 2. linear speed 3. centripetal acceleration of the body.
Answer» Data : m = 100 g = 0.1 kg, r = 50 cm = 0.5 m, ω = 20 rad/s 1. The period of revolution of the body, T = \(\frac{2\pi}{\omega}\) = \(\frac{2\times 3.142}{20}\) = 0.3142 s 2. Linear speed, v = ωr = 20 × 0.5 = 10 m/s 3. Centripetal acceleration, a_c = w²r = (20)² × 0.5 = 200 m/s²

62.

A body of mass 100 grams is tied to one end of a string and revolved along a circular path in the horizontal plane. The radius of the circle is 50 cm. If the body revolves with a constant angular speed of 20 rad/s, find the1. period of revolution 2. linear speed 3. centripetal acceleration of the body.

Answer»

Data : m = 100 g = 0.1 kg, r = 50 cm = 0.5 m, ω = 20 rad/s

1. The period of revolution of the body,

T = \(\frac{2\pi}{\omega}\) = \(\frac{2\times 3.142}{20}\) = 0.3142 s

2. Linear speed, v = ωr = 20 × 0.5 = 10 m/s

3. Centripetal acceleration,

a_c = w²r = (20)² × 0.5 = 200 m/s²

Discussion

63.	A body of mass 100 grams is tied to one end of a string and revolved along a circular path in the horizontal plane. The radius of the circle is 50 cm. If the body revolves with a constant angular speed of 20 rad/s, find the 1. period of revolution 2. linear speed 3. centripetal acceleration of the body.
Answer» Data : m = 100 g = 0.1 kg, r = 50 cm = 0.5 m, ω = 20 rad/s 1. The period of revolution of the body, 2. Linear speed, v = ωr = 20 × 0.5 = 10 m/s 3. Centripetal acceleration, a_c = ω² = (20)² × 0.5 = 200 m/s²

63.

A body of mass 100 grams is tied to one end of a string and revolved along a circular path in the horizontal plane. The radius of the circle is 50 cm. If the body revolves with a constant angular speed of 20 rad/s, find the 1. period of revolution 2. linear speed 3. centripetal acceleration of the body.

Answer»

Data : m = 100 g = 0.1 kg, r = 50 cm = 0.5 m,

ω = 20 rad/s

1. The period of revolution of the body,

2. Linear speed, v = ωr = 20 × 0.5 = 10 m/s

3. Centripetal acceleration,

a_c = ω² = (20)² × 0.5 = 200 m/s²

Discussion

Explore topic-wise InterviewSolutions in .

If L is the angular momentum and I is the moment of inertia of a rotating body, then \(\frac{L^2}{2I}\) represents its (A) rotational PE (B) total energy (C) rotational KE (D) translational KE.

A body, tied to a string, performs circular motion in a vertical plane such that the tension in the string is zero at the highest point. What is the angular speed of the body at the 1. highest position 2. lowest position ?

A body, tied to a string, performs circular motion in a vertical plane such that the tension in the string is zero at the highest point. What is the linear speed of the body at the1. lowest position 2. highest position ?

A bucket of water is whirled in a vertical circle at an arm’s length. Find the minimum speed at the top so that no water spills out. Also find the corresponding angular speed. [Assume r = 0.75 m]

Attach a body of suitable mass to a spring balance so that it stretches by about half its capacity. Now whirl the spring balance so that the body performs a horizontal circular motion. You will notice that the balance now reads more for the same body. Can you explain this ?

A disc of mass 4 kg rolls on a horizontal surface. If its linear speed is 3 m/s, what is its total kinetic energy?

A uniform solid sphere of mass 10 kg rolls on a horizontal surface. If its linear speed is 2 m/s, what is its total kinetic energy?

A spherical shell rolls down a plane inclined at 30° to the horizontal. What is its acceleration ?

Certain stars are believed to be rotating at about 1 rot/s. If such a star has a diameter of 40 km, what is the linear speed of a point on the equator of the star?

An object of mass 0.5 kg is tied to a string and revolved in a horizontal circle of radius 1 m. If the breaking tension of the string is 50 N, what is the maximum speed the object can have?