# Centripetal Acceleration

## Trending Questions

**Q.**

If the earth stops rotating, the apparent value of g on its surface will

Increase everywhere

Decrease everywhere

Remain the same everywhere

Increase at some places and remain the same at some other places

**Q.**

What causes circular motion?

**Q.**The magnitude of the centripetal force acting on a body of mass mexecuting uniform motion in a circle of radius rwith speed vis

- mvr
- mv2r
- vr2m
- vrm

**Q.**

A stone is tied to one end of a string 50 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 10 revolutions in 20 s, what is the magnitude of acceleration of the stone

493cm/s2

720cm/s2

720cm/s2

990cm/s2

**Q.**

A ball of mass 0.1 Kg. is whirled in a horizontal circle of radius 1 m. by means of a string at an initial speed of 10 R.P.M. Keeping the radius constant, the tension in the string is reduced to one quarter of its initial value. The new speed is

5 r.p.m.

10 r.p.m.

20 r.p.m.

14 r.p.m.

**Q.**

If the overbridge is concave instead of being convex, the thrust on the road at the lowest position will be

mg+mv2r

mg−mv2r

m2v2gr

v2gr

**Q.**Two bodies of equal masses revolve in circular orbits of radii R1 and R2 with the same period. Their centripetal forces are in the ratio

- (R2R1)2
- R1R2
- (R1R2)2
- √R1R2

**Q.**

If a particle is performing uniform circular motion with radius R, with angular velocity ⍵, then the magnitude of the acceleration is _____

**Q.**The acceleration of a train travelling with speed of 400 m/s as it goes round a curve of radius 160 m, is

- 1 km/s2
- 100 m/s2
- 10 m/s2
- 1 m/s2

**Q.**A ride in amusement park Imagica consists of rotating circular platform of 6 m in diameter, from which heavy seats are suspended at end of light chains (length of chain =3 m). When the system rotates, chains make an angle of 37∘ with vertical. Find speed of kid sitting on it.

- 6 m/s
- 14 m/s
- 18 m/s
- 12 m/s

**Q.**A 1 kg stone at the end of 1 m long string is whirled in a vertical circle at a constant speed of 4 m/s. The tension in the string is 6 N, when the stone is

(Take g=10 m/s2)

- At the top of the circle
- At the bottom of the circle
- Half way down
- None of above

**Q.**

A body is revolving with a constant speed along a circle. If its direction of motion is reversed but the speed remains the same, then which of the following statement is true

The centripetal force will not suffer any change in magnitude

The centripetal force will have its direction reversed

The centripetal force will not suffer any change in direction

The centripetal force would be doubled

**Q.**A particle is moving in a circular path of radius r with an angular speed ′ω′. Suddenly, the radius is quadrupled and angular speed is halved. Then which of the following statements is correct?

- Centripetal force is doubled
- Centripetal force is halved
- Centripetal force remains same
- None of the above

**Q.**A boy whirls a stone in a horizontal circle at a height of 31.25 m above ground level and radius of circle being 3 m. Somehow, the string breaks and stone flies off horizontally and strikes the ground after travelling horizontal distance of 5 m. Find the magnitude of centripetal acceleration of stone while being in circular motion. Assume stone to be in uniform circular motion.

- 1.75 m/s2
- 6.6 m/s2
- 1.3 m/s2
- 2.6 m/s2

**Q.**If the radius of curvature of the path of two particles of same masses are in the ratio 1:3, then in order to have same centripetal force, their velocity should be in the ratio of

- 1:3
- 3:1
- √3:2
- 1:√3

**Q.**In the given figure, a=15 m/s2 represents the total acceleration of a particle moving in the clockwise direction in a circle of radius R=2.5 m at a given instant of time. The speed of the particle is

- 6.2 m/s
- 4.5 m/s
- 5.0 m/s
- 5.7 m/s

**Q.**For a particle moving on a circular path of radius 5 cm, the velocity at an instant ‘t’ is 3^i+4^j m/s. Assuming the particle to be in uniform circular motion, the value of centripetal acceleration is

- 180 m/s2
- 320 m/s2
- 980 m/s2
- 500 m/s2

**Q.**A table with a smooth horizonal surface is placed in a cabin which moves in a horizontal circle of a large radius R. A smooth pulley of small radius is fixed to the table. Two masses m and 2m placed on the table are connected through a light string going over the pulley. System is released from rest with respect to the cabin. The initial tension in the string will be:

- 13mω2R
- 43mω2R
- 23mω2R
- mω2R

**Q.**A ball of mass 0.9 kg is projected with an initial velocity of 10 m/s at an angle of 37∘ with the horizontal. After 0.6 sec, the gravitational field vanishes and a force of constant magnitude is applied on the ball afterwards, the force being always perpendicular to the direction of motion till it strikes the ground. When it strikes the ground, it is moving vertically downwards. Choose the correct option. (Take g=10 m/s2)

- Path of projectile after 0.6 sec is parabolic
- Path of ball after 0.6 sec will be circular and radius of circle will be 1.8 m
- Magnitude of the constant force applied is 32 N
- Speed during circular motion will be 8 m/s

**Q.**A particle describes a horizontal circle of radius r at a uniform speed, on the inside of the smooth surface of an inverted cone as shown in figure. The height of the plane of the circle above the vertex is h, then the speed of particle should be:

- √rg
- √2rg
- √gh
- √2gh

**Q.**A person with his hands in his pockets is skating on ice at the velocity of 10 m/s and describes a circle of radius 50 m. What is his inclination with vertical

- tan−1(110)
- tan−1(35)
- tan−1(1)
- tan−1(15)

**Q.**A string is passing through a ring of mass m as shown. The ring rotates in a horizontal plane of radius r with a constant angular velocity. What should be the maximum speed of rotation so that the string does not get distorted?

- rg
- √rg
- rg
- gr

**Q.**If the earth goes around the sun in a circular orbit with a constant speed of 30 km/s, the instantaneous acceleration of the earth points towards the sun.

- False
- True

**Q.**

A body of mass 1 kg tied to one end of string is revolved in a horizontal circle of radius 0.1 m with a speed of 3 revolution/sec, assuming the effect of gravity is negligible, then linear velocity, acceleration and tension in the string will be

1.88 m/s, 35.5 m/s2, 35.5 N

2.88 m/s, 45.5 m/s2, 45.5 N

3.88 m/s, 55.5 m/s2, 55.5 N

None of these

**Q.**A cabin rotates on a smooth horizontal table with a uniform angular velocity ω in a circular path of radius R (cabin is in a horizontal plane). A smooth path AB is made inside the cabin as shown in figure. If a particle is released from A to reach B along the path AB, then find the time taken by the particle to reach point B, considering length of path AB =L.

- ln(1+RcosθL)ω
- ln(1+LcosθR)ω√sinθ
- ln(1+LcosθR)ω√cosθ
- ln(1+RcosθL)ω√cosθ

**Q.**Two protons of masses 1.6×10−27 kg, goes round in a concentric circular orbits of radii 0.50 nm and 1 nm experience centripetal forces in the ratio of 1:4, then the ratio of frequency of revolution of the protons is (Assume no force of interaction between protons )

- 1:2
- 2:1
- 1:√2
- √3:1

**Q.**A pilot of mass 75 kg in a jet aircraft executes a loop with a constant speed of 360 km/hr. If the radius of the loop is 200 m, the force exerted by seat on the pilot at the top of the loop is (Take g=10 m/s2)

- 3750 N
- 4485 N
- 3000 N
- 5843 N

**Q.**

If a cycle wheel of radius 4 m completes one revolution in two seconds. Then acceleration of a point on the cycle wheel will be

π2m/s2

2π2m/s2

4π2m/s2

8πm/s2

**Q.**

A mass of 2 kg is whirled in a horizontal circle by means of a string at an initial speed of 5 revolutions per minute. Keeping the radius constant the tension in the string is doubled. The new speed is nearly

14 rpm

10 rpm

2.25 rpm

7 rpm

**Q.**A particle moves with deceleration along the circle of radius R so that at any moment of time its tangential and centripetal accelerations are equal in moduli. At the initial moment t=0, the speed of the particle equals v0. Then

The speed of the particle as a function of the distance covered s will be

- v=v0e−sR
- v=v0esR
- v=v0e−R/s
- v=v0eR/s