Circular Motion

Q. A ball is swung in a vertical circle. At the highest point of the circle, what is the condition for the ball to just maintain its circular motion?

A. Weight must be greater than tension
B. Tension must be zero
C. Centripetal force must be zero
D. Weight must be less than tension

Solution

At the highest point, the centripetal force is provided by the weight of the ball. For just maintaining motion, tension can be zero.

Correct Answer: B — Tension must be zero

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Q. A ball is tied to a string and swung in a vertical circle. At the highest point of the circle, what is the condition for the ball to just maintain circular motion?

A. Tension = 0
B. Tension = mg
C. Tension > mg
D. Tension < mg

Solution

At the highest point, the centripetal force is provided by the weight, so T + mg = mv²/r, T = 0.

Correct Answer: A — Tension = 0

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Q. A ball is tied to a string and swung in a vertical circle. At the highest point of the circle, what is the condition for the ball to remain in circular motion?

A. Tension must be zero
B. Tension must be maximum
C. Weight must be zero
D. Centripetal force must be zero

Solution

At the highest point, the tension can be zero if the centripetal force is provided entirely by the weight.

Correct Answer: A — Tension must be zero

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Q. A ball is tied to a string and swung in a vertical circle. At the highest point, the tension in the string is 2 N and the weight of the ball is 3 N. What is the speed of the ball at the highest point if the radius of the circle is 1 m?

A. 1 m/s
B. 2 m/s
C. 3 m/s
D. 4 m/s

Solution

At the highest point, T + mg = mv²/r. 2 N + 3 N = mv²/1. v² = 5, v = √5 ≈ 2.24 m/s.

Correct Answer: D — 4 m/s

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Q. A body is moving in a circular path of radius 4 m with a constant speed of 8 m/s. What is the net force acting on the body if its mass is 2 kg?

A. 4 N
B. 8 N
C. 16 N
D. 32 N

Solution

Centripetal force F = m(v²/r) = 2(8²/4) = 32 N.

Correct Answer: C — 16 N

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Q. A car is moving in a circular path of radius 50 m with a constant speed of 20 m/s. What is the centripetal acceleration of the car?

A. 2 m/s²
B. 4 m/s²
C. 8 m/s²
D. 10 m/s²

Solution

Centripetal acceleration (a_c) = v²/r = (20 m/s)² / (50 m) = 8 m/s².

Correct Answer: B — 4 m/s²

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Q. A car is moving in a circular path of radius 50 m with a speed of 15 m/s. What is the angular displacement after 10 seconds?

A. 1 rad
B. 2 rad
C. 3 rad
D. 4 rad

Solution

Angular displacement θ = (v/r)t = (15/50)(10) = 3 rad.

Correct Answer: D — 4 rad

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Q. A car is moving in a circular track of radius 100 m at a speed of 20 m/s. What is the time period of one complete revolution?

A. 10 s
B. 20 s
C. 30 s
D. 40 s

Solution

Circumference = 2πr = 2π(100) = 200π m. Time period (T) = Circumference / Speed = 200π / 20 = 10π s.

Correct Answer: B — 20 s

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Q. A car is moving in a circular track of radius 50 m with a speed of 15 m/s. What is the angular momentum of the car if its mass is 1000 kg? (2000)

A. 7500 kg m²/s
B. 10000 kg m²/s
C. 15000 kg m²/s
D. 20000 kg m²/s

Solution

Angular momentum L = mvr = 1000 * 15 * 50 = 750000 kg m²/s.

Correct Answer: A — 7500 kg m²/s

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Q. A car is negotiating a curve of radius 100 m at a speed of 15 m/s. What is the minimum coefficient of friction required to prevent the car from skidding?

A. 0.15
B. 0.25
C. 0.30
D. 0.35

Solution

Frictional force = m * a_c; μmg = mv²/r; μ = v²/(rg) = (15 m/s)² / (100 m * 9.8 m/s²) ≈ 0.23.

Correct Answer: B — 0.25

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Q. A car travels around a circular track of radius 50 m at a speed of 15 m/s. What is the centripetal force acting on the car if its mass is 1000 kg?

A. 450 N
B. 225 N
C. 150 N
D. 75 N

Solution

Centripetal force (F_c) = mv²/r = 1000 kg * (15 m/s)² / 50 m = 4500 N.

Correct Answer: A — 450 N

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Q. A conical pendulum consists of a mass attached to a string that swings in a horizontal circle. If the angle of the string with the vertical is θ, what is the expression for the tension in the string?

A. mg/cos(θ)
B. mg/sin(θ)
C. mg/tan(θ)
D. mg

Solution

Tension T = mg/cos(θ) to balance the vertical component of weight.

Correct Answer: A — mg/cos(θ)

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Q. A conical pendulum consists of a mass m attached to a string of length L, swinging in a horizontal circle. What is the expression for the tension in the string?

A. T = mg
B. T = mg/cos(θ)
C. T = mg/sin(θ)
D. T = m(v²/r)

Solution

In a conical pendulum, T = mg/cos(θ) where θ is the angle with the vertical.

Correct Answer: B — T = mg/cos(θ)

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Q. A conical pendulum swings in a horizontal circle. If the angle of the string with the vertical is θ, what is the relationship between the tension in the string and the gravitational force acting on the pendulum bob?

A. T = mg
B. T = mg cos(θ)
C. T = mg sin(θ)
D. T = mg tan(θ)

Solution

Tension provides the vertical component to balance the weight: T cos(θ) = mg.

Correct Answer: B — T = mg cos(θ)

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Q. A conical pendulum swings in a horizontal circle. If the angle of the string with the vertical is θ, what is the relationship between the tension and the gravitational force acting on the pendulum bob?

A. T = mg
B. T = mg cos(θ)
C. T = mg sin(θ)
D. T = mg tan(θ)

Solution

The vertical component of tension balances the weight: T cos(θ) = mg.

Correct Answer: B — T = mg cos(θ)

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Q. A conical pendulum swings in a horizontal circle. If the angle of the string with the vertical increases, what happens to the tension in the string?

A. Increases
B. Decreases
C. Remains the same
D. Becomes zero

Solution

As the angle increases, the vertical component of tension must increase to balance the weight.

Correct Answer: A — Increases

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Q. A conical pendulum swings in a horizontal circle. If the angle of the string with the vertical is θ, what is the expression for the tension in the string?

A. T = mg
B. T = mg/cos(θ)
C. T = mg/sin(θ)
D. T = mg tan(θ)

Solution

The vertical component of tension balances the weight: T cos(θ) = mg, thus T = mg/cos(θ).

Correct Answer: B — T = mg/cos(θ)

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Q. A conical pendulum swings in a horizontal circle. If the angle of the string with the vertical is θ, what is the relationship between the tension in the string and the gravitational force?

A. T = mg
B. T = mg/cos(θ)
C. T = mg/sin(θ)
D. T = mg/tan(θ)

Solution

Tension T provides the centripetal force and balances the weight, T = mg/cos(θ).

Correct Answer: B — T = mg/cos(θ)

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Q. A conical pendulum swings with a constant speed. If the angle of the string with the vertical is θ, what is the expression for the tension in the string?

A. mg/cos(θ)
B. mg/sin(θ)
C. mg/tan(θ)
D. mg

Solution

Tension T = mg/cos(θ) to balance the vertical component of weight.

Correct Answer: A — mg/cos(θ)

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Q. A cyclist is moving around a circular track of radius 100 m. If he completes one lap in 40 seconds, what is his average speed?

A. 5 m/s
B. 10 m/s
C. 15 m/s
D. 20 m/s

Solution

Circumference = 2πr = 2π(100 m). Average speed = distance/time = (2π(100 m))/40 s ≈ 15.7 m/s.

Correct Answer: B — 10 m/s

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Q. A cyclist is moving in a circular path of radius 10 m at a speed of 5 m/s. What is the angle of banking required to prevent slipping?

A. 30°
B. 45°
C. 60°
D. 90°

Solution

tan(θ) = v²/(rg) = (5²)/(10*9.8) = 0.25. θ = tan⁻¹(0.25) ≈ 14°.

Correct Answer: A — 30°

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Q. A cyclist is moving in a circular path of radius 15 m with a speed of 6 m/s. What is the angular velocity of the cyclist?

A. 0.4 rad/s
B. 0.6 rad/s
C. 0.8 rad/s
D. 1.0 rad/s

Solution

Angular velocity (ω) = v/r = 6 m/s / 15 m = 0.4 rad/s.

Correct Answer: B — 0.6 rad/s

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Q. A cyclist is moving in a circular track of radius 30 m with a speed of 15 m/s. What is the net force acting on the cyclist if the mass of the cyclist is 60 kg?

A. 180 N
B. 120 N
C. 90 N
D. 60 N

Solution

Centripetal force F = mv²/r = 60 kg * (15 m/s)² / 30 m = 180 N.

Correct Answer: A — 180 N

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Q. A cyclist is moving in a circular track of radius 30 m. If he completes one round in 12 seconds, what is his average speed?

A. 5 m/s
B. 10 m/s
C. 15 m/s
D. 20 m/s

Solution

Circumference = 2πr = 2π(30) = 60π m. Average speed = distance/time = 60π m / 12 s = 5 m/s.

Correct Answer: B — 10 m/s

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Q. A cyclist is moving in a circular track of radius 30 m. If the cyclist completes one round in 12 seconds, what is the average speed of the cyclist?

A. 5 m/s
B. 10 m/s
C. 15 m/s
D. 20 m/s

Solution

Circumference = 2πr = 2π(30 m). Average speed = distance/time = (2π(30 m))/12 s ≈ 5 m/s.

Correct Answer: B — 10 m/s

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Q. A cyclist is moving in a circular track of radius 30 m. If the cyclist completes one round in 12 seconds, what is the angular velocity of the cyclist?

A. π/6 rad/s
B. π/3 rad/s
C. 2π/6 rad/s
D. 2π/3 rad/s

Solution

Angular velocity (ω) = 2π/T = 2π/12 = π/6 rad/s.

Correct Answer: B — π/3 rad/s

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Q. A cyclist is negotiating a circular track of radius 30 m. If the cyclist's speed is 15 m/s, what is the net force acting on the cyclist towards the center of the track?

A. 50 N
B. 75 N
C. 100 N
D. 125 N

Solution

Centripetal force (F_c) = mv²/r. Assuming mass m = 100 kg, F_c = (100 kg)(15 m/s)² / (30 m) = 75 N.

Correct Answer: C — 100 N

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Q. A cyclist is negotiating a circular turn of radius 30 m at a speed of 15 m/s. What is the minimum coefficient of friction required to prevent slipping?

A. 0.25
B. 0.5
C. 0.75
D. 1

Solution

Frictional force = m * a_c; μmg = mv²/r; μ = v²/(rg) = (15²)/(30*9.8) = 0.25.

Correct Answer: B — 0.5

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Q. A mass m is attached to a string and is whirled in a horizontal circle. If the radius of the circle is halved, what happens to the tension in the string if the speed remains constant?

A. It doubles
B. It remains the same
C. It halves
D. It quadruples

Solution

Tension T = mv²/r. If r is halved, T doubles for constant speed.

Correct Answer: A — It doubles

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Q. A mass m is attached to a string and is whirled in a vertical circle. At the highest point of the circle, what is the condition for the mass to just complete the circular motion?

A. Tension = 0
B. Tension = mg
C. Tension = 2mg
D. Tension = mg/2

Solution

At the highest point, the centripetal force is provided by the weight of the mass, so T + mg = mv²/r. For T = 0, mg = mv²/r.

Correct Answer: A — Tension = 0

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Circular Motion

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