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#### Example – 01:

- A turntable rotates at 100 rev/sec. Calculate its angular speed in rad/s and degrees/s.
- Solution:
**Given:**n = 100 r.p.s.**To Find:**Angular speed =ω =?

**Ans:** Angular speed = 10.47 rad/s, Angular speed = 600 degrees/s

#### Example – 02:

- An object of mass 100 g moves around the circumference of a circle of radius 2m with a constant angular speed of 7.5 rad/s. Compute its linear speed and force directed towards centre.
**Solution:****Given:**mass of thre body = m = 100 g = 0.1 kg, Radius of circular path = r = 2 m, Angulae speed = ω = 7.5 rad/s,**To find:**Linear speed = v = ?, Centripetal force = F = ?

v = r ω = 2 x 7.5 = 15 m/s

F = m r ω^{2}

∴ F = 0.1 x 2 x (7.5)^{2 }= 11.25 N

**Ans: **Linear speed = 15 m/s, Centripetal force = 11.25 N radially inward.

#### Example – 03:

- A spherical bob of diameter 3 cm having a mass 100 g is attached to the end of a string of length 48.5 cm. Find the angular velocity and the tension in the string, if the bob is rotated at a speed of 600 r.p.m. in a horizontal circle. If the same bob is now whirled in a vertical circle of the same radius, what will be the difference in the tensions at the lowest point and the highest point?
**Solution:****Given:**Mass of bob = m = 100 g = 0.1 kg, Radius of circular path = r = 48.5 cm + 1.5 cm = 50 cm = 0.5 m, rpm = N = 600 r.p.m.,**To Find:**Angular velocity = ω= ?, Tension in the string = F = ?

The necessary centripetal force acting on a body is given by tension in the string

The difference in the tensions at the lowest point and the highest point

= 6mg = 6 x 0.1 x 9.8 = 5.88 N

**Ans: **Angular speed = 62.84 rad/s, The tension in string =.197.2 N

The difference in the tensions at the lowest point and the highest point is 5.88 N

#### Example – 04:

- A flat curve highway has a radius of curvature 400 m. A car rounds the curve at a speed of 32 m/s. What is the minimum value of the coefficient of friction that will prevent the car from sliding?
**Given:**Radius of curvature = r = 400 m, speed of car = 32 m/s, g = 9.8 m/s^{2}.**To Find:**Coefficient of Friction = m =?**Solution:**

The necessary centripetal force is provided by the friction between the road and tyres of the car

**Ans:** The coefficient of friction is 0.2612

#### Example – 05:

- A metre gauge train is moving at 60 km/hr along a curved rod of a radius of curvature 500m at a certain place. Find the elevation of outer rail over the inner rail, so that there is no side pressure on the rail. (g = 9.8 m/s
^{2}) Ans: (0.0567 m.) **Given:**Speed of train = 60 kmph = 60 x 5/18 =16.67 m/s, radius of curve = r = 500 m, distance between rail =*l*= 1 m. g = 9.8 m/s^{2}**To find:**elevation of outer rail over the inner rail = h =?**Solution:**

The angle of banking is given by

Elevation h =* l* sinθ = 1 x sin (3^{o}15’) = 1x 0.0567 =0.0567 m = 5.67 cm

**Ans:** The angle of banking is 3^{o}15’ and the elevation of outer rail over the inner rail is 5.67 cm

#### Example – 06:

- An aircraft in level flight completes a circular turn in 100 seconds. What is the radius of the circular turn? What is the angle of banking, if the velocity of aircraft is 40 m/s?
**Given:**Time taken = T = 100 s. velocity of aircraft = 40 m/s.**To Find:**radius of circular turn = ?, angle of banking = θ = ?**Solution:**

100 seconds are taken to complete a circular turn (circumference)

**Ans:** The angle of banking of aircraft is 14°25′

#### Example – 07:

- In a circus, motorcyclist having mass 50 kg move in a spherical cage of radius 3m. Calculate the least velocity with which he must pass the highest point without loosing. Also, calculate the angular speed at the highest point.
**Solution:****Given:**Radius of hollow sphere = r = 3 m,**To find:**Minimum speed required = v = ?, Angular speed = ω = ?

At the highest point, the centrifugal force and weight of car are equal in magnitude.

**Ans : **The velocity of car = 5.42 m/s, Angular speed = 1.807 rad/s.

#### Example – 08:

- A coin is placed at a distance of 10 cm from the centre of a turntable of radius 1m just begins to slip, when the turntable rotates at a speed of 90 r.p.m. Calculate the coefficient of static friction between the coin and the turntable.
**Given:**Radius of circle = r = 10 cm = 0.1 m, speed of rotation = N = 90 r.p.m.**To Find:**Coefficient of Friction = m =?**Solution:**

The necessary centripetal force is provided by the friction between the turntable and coin

Ans: The coefficient of friction is 0.9066

**Example – 09:**

- A stone is whirled in a vertical circle at the end of a rope of length 0.5 m. Find the velocity of a stone at a) lowest position b) midway when the string is horizontal c) topmost position to just complete the circle.
**Solution:****Given:**Radius of circular path = r = 0.5 m, Mass of the body = m = 1 kg, g = 9.8 m/s^{2},**To find:**velocity at lowest point = v_{L}= ?, velocity when string is horizontal = v_{M}= ?, velocity at topmost point = v_{H}= ?.

**Ans:** Velocity of stone at the lowermost point = 4.95 m/s,

The velocity of stone when the string is horizontal = 3.83 m/s,

The velocity of stone at the topmost point = 2.21 m/s

#### Example – 10:

- A stone of mass 1 kg is whirled in a horizontal circle attached at the end of 1 m long string making an angle of 30° with the vertical. Find the period and centripetal force if g = 9.8m/s
^{2}. **Solution:****Given:**Length of pendulum = l = 1 m, angle with vertical = θ = 30°, g = 9.8 m/s^{2},**To find:**Period = T = ?, Centripetal force = F = ?

**Ans: **Period of motion = 1.867 s, Centripetal force = 5.657 N

#### Example – 11:

- A wheel has radius 30 cm. The wheel starts from rest and attains a speed of 300 r.p.m. in 3 minutes. What is angular acceleration? What is angular displacement during this time?
**Given:**Radis of wheel = 30 cm = 0.3 m, Initial angular speed = N_{1}= 0 r.p.m., Final angular speed = N_{2}= 300 r.p.m.,- Time taken = 3 minutes = 3 x 60 = 180 s
**To Find:**Angular Acceleration a = ?, Angular displacement = θ = ?**Solution:**

**Ans:** The angular acceleration is 0.1745 rad/s^{2} and angular displacement is 2827 rad

**Example – 12:**

- A rotor has a diameter of 4.0 m. The rotor is rotated about the central vertical axis. The occupant remains pinned against the wall. When the linear velocity of the drum is 8 m/s. Compute the coefficient of static friction between the wall of the rotor and the clothing of occupant. Also, calculate the angular velocity of the drum. How many revolutions will it make in a minute?
**Solution:****Given:**diameter = d = 4 m, radius = r = 4/2 = 2 m, linear speed = v =.8 m/s , g = 9.8 m/s^{2},**To find:**Coefficient of friction = μ = ?, Angular velocity = ω = ?, rpm = N = ?

** **

*As the occupant remains pinned against the wall, his weight is equal to the frictional force.*

*Frictional force = Weight of a body*

**Ans: **Coefficient of friction = 0.3063, Angular velocity = 4 rad/s, Number of revolutions per minute = 38.21.

Science > Physics > Circular Motion > You are Here |

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