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Maharashtra State Board > Science > Physics Question Bank > Circular Motion > You are Here |

**Very Short Answers (1 Mark)**

**Q 1. What is angular displacement in a radian of the second hand of a clock in 10 seconds?**

**Given: **For second hand T = 60 sec, t = 10 sec, **To Find: **Angular displacement θ =?

**Solution: **θ = ω t = (2π/T) t = (2π/60) x 10 = π /3 radian

Ans: Thus displacement of the second hand in 10 seconds is π /3 radian

**Q2. Write vector relation between angular velocity (ω), tangential velocity (v) and position vector (r).**

The required relation is

**Q3. A cycle wheel is rotating with uniform angular velocity (ω). What is the nature of the graph between tangential velocities of different points on a spoke and their distances from the axis?**

- As angular velocity is constant, the tangential velocity of a point is directly proportional to the distance of the point from the centre of the wheel. The graph is a straight line passing through the origin.

**Q4. What is the ratio of angular velocities of hour – hand of a clock and the spin motion of the earth?**

**Ans: **The ratio of the angular velocities of hour – hand of a clock and the spin motion of the earth is 2: 1

**Q5. A body of mass ‘M’ is revolving in a vertical circle of radius ‘r’. What is the difference in the kinetic energies at the bottom and top position of the circle?**

Kinetic energy at the bottom position of vertical circle = E_{L} = (5/2)Mgr

Kinetic energy at the top position of vertical circle = E_{T} = (1/2)Mgr

Difference in kinetic energies at the two positions = E_{L} – E_{T} = (5/2)Mgr – (1/2)Mgr = 2 Mgr

Where M = mass of the body

g = acceleration due to gravity

r = radius of the circular path

**Q6. Why work done by the centripetal force is zero?**

- The work done by a force is defined as the product of force and the displacement in the direction of the force.
- The work done by the centripetal force is zero because the displacement of the particle (tangential) is perpendicular to the direction of the centripetal force (radial) i.e. there is no displacement in the direction of the centripetal force.

W = F s = F(0) = 0

**Q7. What is the effect of centripetal force on earth at the equator and at poles?**

- As we move from the equator towards the pole the distance of the point on the surface of the earth from the axis of rotation decreases. Thus the centripetal force due to rotation of the earth acting on a body is maximum at equator and minimum (zero) at poles.
- Thus the net force acting on body increases from the equator towards the pole and hence the weight of the body is minimum at equator and maximum at poles.

### Short Answers – I (2 Marks)

**Q1. Write S. I. unit of angular velocity. State the rule concerned with the direction of angular velocity.**

- The S.I. unit of angular velocity is rad/s (radians per second). The direction of the angular velocity vector is given by the right-hand thumb rule.
**Right-Hand Thumb Rule:**Imagine the axis of rotation to be held in the right hand with the fingers curled around it and the thumb outstretched. If the curled fingers give the direction of motion of a particle performing the circular motion, then the direction of outstretched thumb gives the direction of the angular velocity vector.

**Q2. Obtain the relation between linear velocity and angular velocity.**

- Consider a particle performing uniform circular motion, along the circumference of the circle of radius ‘r’ with constant linear velocity ‘v’ and constant angular speed ‘ω’ moving in the anticlockwise sense as shown in the figure.
- Suppose the particle moves from point P to point Q through a distance ‘δx’along the circumference of the circular path and subtends the angle ‘δθ’ at the centre O of the circle in a small interval of time ‘δt’. By geometry

δx = r . δθ

- If the time interval is very very small then arc PQ can be considered to be almost a straight line. Therefore the magnitude of linear velocity is given by

- Thus the linear velocity of a particle performing uniform circular motion is radius times its angular velocity. In vector form above equation can be written as

**Q3. Which physical quantities remain constant in U.C.M.?**

- In U.C.M. the linear speed, angular velocity, angular acceleration (zero), period, frequency, centripetal acceleration, centripetal force, angular momentum and kinetic energy remain constant.

**Q4. Define the centrifugal force. Explain its any one example.**

- The imaginary (pseudo) force which acts on the particle performing a circular motion in the direction away from the centre along the radius of the circular path having the same magnitude as that of centripetal force is called as centrifugal force.
**Example:**When moving car takes a turn along a horizontal curved road, persons in the car experience a force in the outward direction. This force is centrifugal force

**Q5. A particle is performing U.C.M. along a circle of radius r. In half period of revolution, what is its displacement and corresponding distance?**

- In half period the particle will be at the diametrically opposite position to original position. Hence the displacement is 2r and corresponding distance travelled is half the circumference = (1/2) 2πr = πr.

**Short Answers II (3 Marks)**

**Q1. Define angular velocity and angular acceleration. Obtain relation between linear velocity and angular velocity.**

**Angular Velocity:**The angular velocity of a particle performing circular motion is defined as the time rate of change of limiting angular displacement.**Angular Acceleration:**The average angular acceleration is defined as the time rate of change of angular velocity.

- Consider a particle performing uniform circular motion, along the circumference of the circle of radius ‘r’ with constant linear velocity ‘v’ and constant angular speed ‘ω’ moving in the anticlockwise sense as shown in the figure.
- Suppose the particle moves from point P to point Q through a distance ‘δx’along the circumference of the circular path and subtends the angle ‘δθ’ at the centre O of the circle in a small interval of time ‘δt’. By geometry

δx = r . δθ

- If the time interval is very very small then arc PQ can be considered to be almost a straight line. Therefore the magnitude of linear velocity is given by

- Thus the linear velocity of a particle performing uniform circular motion is radius times its angular velocity. In vector form above equation can be written as

**Q2. Obtain the vector relation between linear velocity and angular velocity. Show that v ∝ n. where ‘n’ is the frequency of revolution.**

- For smaller magnitudes angular displacement, angular velocity are vector quantities. Let ( r) be the position vector of the particle at some instant. Let the angular displacement in small time δt be (dq). Let the corresponding linear displacement (arc length) be ( ds). By geometry

Dividing both sides of the equation by δt and taking the limit

Considering magnitude only

v = r ω = r (2π n)

v = (2πr) n

The quantities in the bracket are constant

v ∝ n (proved as required)

**Q3. Show that acceleration in U.C.M. is v²/r**

- The magnitude of the velocity of a particle performing uniform circular motion is constant but its direction changes constantly in the direction. Hence the particle in circular motion has linear acceleration.
- Let us consider a particle performing uniform circular motion with a linear velocity of magnitude v and angular velocity of magnitude ω along a circle of radius r with centre O in an anticlockwise sense (moving from initial position A to final position B)as shown in the figure.

AP = velocity of the particle at position A

BQ = velocity of the particle at position B

**Construction:**Draw BR || AP and BR =| AP and join RQ

Consider velocity triangle QBR. By triangle law of vector addition

Now by definition,

The triangles AOB and RBQ are similar. Hence ∠ QBR = δθ

For smaller angular displacement δθ,

Substituting in equation (1), we get

This is the expression for acceleration of particle performing the uniform circular motion.

- This acceleration is directed towards the centre of the circular path along the radius. This acceleration is called a radial acceleration or centripetal acceleration.

**4. Define conical pendulum. Prove that T ∝ √ l where ‘ l ‘ is the length of string and T is periodic time.**

- Let us consider a conical pendulum consists of a bob of mass, ‘m’ revolving in a horizontal circle with constant speed ‘v’ at the end of a string of length ‘l’. Let the string makes a constant angle ‘θ’ with the vertical. let ‘h’ be the depth of the bob below the support.

- The tension ‘F’ in the string can be resolved into two components. Horizontal ‘Fsin θ’ and vertical ‘Fcos θ’.

The vertical component (F cos θ) balances the weight mg of the vehicle.

F cosθ = mg ………….. (1)

The horizontal component (F sin θ) provides the necessary centripetal force.

F sin θ = mv^{2}/r ………… (2)

Dividing equation (2) by (1) we get,

But v = rω

Where ω is angular speed and T is the period of the pendulum.

From figure tan θ = r/h

Substituting in equation (4)

This is an expression for the time period of a conical pendulum.

The quantities 2, π, θ, and g are constant.

T ∝ √*l * (Proved as required)

**Q5. In a vertical circular motion, at which point tension is maximum? Obtain an expression for it at a point midway between the path of vertical circular motion.**

- The tension is maximum at the lowermost point in the vertical circle.
- Consider a small body of mass, ‘m’ attached to one end of a string and whirled in a vertical circle of radius ‘r’. In this case, the acceleration of the body increases as it goes down the vertical circle and decreases when goes up the vertical circle. Hence the speed of the body changes continuously. It is maximum at the bottommost position and minimum at the uppermost position of the vertical circle. Hence the motion of the body is not uniform circular motion. Irrespective of the position of the particle on the circle, the weight ‘mg’ always acts vertically downward.

- Let ‘v’ be the velocity of the body at any point P on the vertical circle. Let L be the lowest point of the vertical circle. Let ‘h’ be the height of point P above point L. let ‘u’ be the velocity of the body at L. By the law of conservation of energy

Energy at point P = Energy at point L

This is an expression for the velocity of a particle at any point performing a circular motion in a vertical circle.

Consider the centripetal force at point P

Substituting in equation (2)

This is the expression for the tension in the string.

at a point midway between the path of vertical circular motion

When the string is horizontal, i.e. the body is at M (h = r)

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