Centripetal Acceleration Questions

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Very Short Answers Questions [1 Mark Questions]

Ques. Define centripetal acceleration. 

Ans. The acceleration produced in a body moving in a circular path due to the continuous change in the direction of motion is called centripetal acceleration. The direction of centripetal acceleration is towards the center of the circular path.

Ques. Acceleration is defined as the change in _____ per unit time.

1. Displacement
2. Velocity
3. Distance
4. None of the above

Ans. The correct answer is b. Velocity

Explanation: The rate of change of velocity is known as acceleration.

Ques. What is the unit of centripetal acceleration?

1. m²/s
2. m²/s²
3. m/s²
4. m/s

Ans. The correct answer is c. m/s²

Explanation: The unit of centripetal acceleration is meter-per-square-second (m/s²).

Ques. For a body moving in a circular path, the centripetal force always acts toward the…………..of the circular path.

1. Sides
2. Center
3. Edges
4. None of the above

Ans. The correct answer is b. Center

Explanation: Centripetal force always acts towards the center of a circular path.

Ques. The formula of centripetal force is given by

1. F = mv² + r²
2. F = mv/r²
3. F = mv²/r
4. F= mv²/r²

Ans. The correct answer is c. F = mv²/r

Explanation: The magnitude of centripetal force acting on a body moving in a circular path is equal to the product of the mass of the body and its centripetal acceleration. I.e

F = mɑ = mv²/r

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Short Answers Questions [2 Marks Questions]

Ques. A plane is circulating around a path of radius 5 km at a constant speed of 15 km/s. What is the centripetal acceleration?

Ans. Given

• The radius of the circular path, r = 5 km
• Speed of the plane, v = 15 km/s

The centripetal acceleration is given by

a_C = v²/r

⇒ a_C = 15²/5 = 45 m/s²

Ques. What is the dimensional formula of centripetal acceleration?

Ans. The formula for centripetal acceleration is given by

a_C = v²/r

Where

• v is velocity
• r is the radius of the circular track

The dimensions of velocity are [v] = [ M⁰ L T^-1 ]

The dimensions of the radius is [r] = [ M⁰ L T⁰ ]

Therefore the dimensional formula of centripetal acceleration is

[a_C] = [v²]/[r] = [ M⁰ L² T^-2 ] / [ M⁰ L T⁰ ]

⇒ [a_C] = [ M⁰ L T^-2 ] = [ L T^-2 ]

Ques. How does centripetal acceleration differ from linear acceleration?

Ans. Centripetal acceleration is defined as the rate of change of tangential velocity i.e. the velocity of an object measured at any point tangent to a circle. 

Linear acceleration is the rate of change of linear velocity i.e. the velocity of an object traveling in a straight line.

Ques. Define centrifugal force.

Ans. A force that tends to move the bodies or particles of a rotating frame, away from the axis of rotation is called centrifugal force. The magnitude of the centrifugal force is equal to the magnitude of the centripetal force.

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Long Answers Questions [3 Marks Questions]

Ques. A 30 kg boy is riding a merry-go-round with a radius of 2 m. What is the centripetal force on the boy if his velocity is 6 m/s?

Ans. Given

• Mass of the boy, m = 30 kg
• The radius of the circular path, r = 2 m
• Velocity, v = 6 m/s

The centripetal force is given by

F_C = mv²/r

⇒ F_C = (30 x 6²)/2 = 540 N

Ques. A boy of mass 35 kg is cycling along a circular path of radius 6 m with a velocity of 2 m/s. Calculate the centripetal acceleration of the boy.

Ans. Given

• Mass of the boy, m = 35 kg
• The radius of the circular path, r = 6 m
• Velocity, v = 2 m/s

The centripetal acceleration is given by

a_C = v²/r

⇒ a_C = 6²/2 = 18 m/s²

Ques. A car of mass m is driving around a circular track of radius r at a constant velocity of v. The centripetal force acting on the car is FC. If the velocity of the car is doubled, what is the new centripetal force required for the car to drive on the circular track?

Ans. Initially for the car of mass m moving with velocity v in a circular track of radius r, the centripetal force is given by

F_C = mv²/r

If the velocity of the car gets doubled then, the velocity will be

v_N = 2v

The new value of centripetal force is given by

F_N = mvN²/r = m(2v)²/r = 4mv²/r = 4FC

Ques. A man of mass 60 kg cycling along a circular track with a speed of 1 m/s. If the centripetal force acting on the man is 20 N. What will the radius of the circular track be?

Ans. Given

• Mass of the man, m = 60 kg
• Centripetal force, F_C = 20 N
• Velocity, v = 1 m/s

The centripetal force is given by

F_C = mv²/r

⇒ r = mv²/F_C

Where r is the radius of the circular track

⇒ r = (60 x 1²)/20 = 3 m

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Very Long Answers Questions [5 Marks Questions]

Ques. A ball is attached to a string that is 2.5 m long. It is spun so that it completes two full rotations every second. What is the centripetal acceleration felt by the ball?

Ans. When the ball is attached to the string rotates it will move in a circular path of radius equal to the length of the string. Therefore

The radius of the circular path, r = 2.5 m

Now it is given that the ball completes two rotations per second, therefore 

The frequency of the rotation, f = 2 s^-1

Angular frequency of the ball, ω = 2πf = 2π x 2 = 4π rad/s

The velocity of the ball is given by

v = ωr = 4π x 2.5 = 10π m/s

The centripetal acceleration of the ball is given by

a = v²/r = (10π)²/2.5 = 395 m/s²

Ques. A car starts from rest on a horizontal circular road of radius 190 m and gains speed at a uniform rate of 1.2 m/s². The coefficient of static friction between the road and the tyres is 0.37. Calculate the distance traveled by the car before it begins to skid.

Ans. When a car is moving in a non-uniform circular motion, then two types of force are acting on the car. These are tangential force (Ft) and centripetal force (FC).

The resultant of both forces is

F_net = √(F_C² + F_t²) = √[(ma_C)² + (ma_t)²] = m√(a_C² +a_t²) 

But centripetal accleration, a_C = v²/R

⇒ F_net = m√(v⁴/R² + at²)

The maximum force of friction acting between the car and the road is

f_max = µ_sN = µ_s mg = F_net

⇒ µ_s mg = m√(v⁴/R² + a_t²)

Squaring both sides, we get

µ_s² m²g² = m² (v⁴/R² + a_t²)

⇒ v⁴ = R² (µ_s²g² - a_t²)

⇒ v_max = [R² (µ_s²g² – a_t²)]¼

Now, we have u = 0, v = [R² (µ_s²g² - a_t²)]¼ and a_t = 1.2 m/s²

Using the equation of motion, v² = u² + 2aS

⇒ S = (v² - u²)/2a

On substituting the values of u, v, and a, we get

Distance traveled by the car, S = 270 m

Ques. A body weighing 0.4 kg is whirled in a verticle circle making 2 revolutions per second. If the radius of the circle is 1.2 m, find the tension in the string when the body is

1. At the bottom of the circle
2. At the top of the circle

Ans. Given

• Mass, m = 0.4 kg
• The radius of the circle, r = 1.2 m
• Frequency of the revolution, f = 2 rps

The angular frequency of the revolution, ω = 2πf = 2π x 2 = 12.56 rad/s

Linear velocity, v = rω = 1.2 x 12.56 = 15.072 m/s

1. At bottom

T_B - mg = ma_C

Where T_B is the tension on the string when the body is at bottom

⇒ T_B = mg + ma_C

⇒ T_B = mg + mv²/r = (0.4 x 9.8) + (0.4 x 15.0722)/1.2

⇒ T_B = 79.64 N

2. At top

T_H + mg = ma_C

Where T_H is the tension on the string when the body is at top

⇒ T_H = ma_C - mg

⇒ T_B = mv²/r - mg = (0.4 x 15.0722)/1.2 - (0.4 x 9.8)

⇒ T_B = 71.8 N

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