Advertisements
Advertisements
Question
Answer the following question.
Show that the centripetal force on a particle undergoing uniform circular motion is -mrω2.
Advertisements
Solution
- Suppose a particle is performing U.C.M in anticlockwise direction. The co-ordinate axes are chosen as shown in the figure.
Let, A = initial position of the particle which lies on positive X-axis
P = instantaneous position after time t
θ = angle made by radius vector
ω = constant angular speed
`vec"r"` = instantaneous position vector at time t - From the figure,
`vec"r" = hat"i""x" + hat"j""y"`
where, `hat"i" and hat"j"` are unit vectors along X-axis and Y-axis respectively.
- Also, x = r cos θ and y = r sin θ
∴ `vec"r" = ["r"hat"i"cos theta + "r"hat"j"sin theta]`
But θ = ωt
∴ `vec"r" = ["r"hat"i" cos omega"t" + "r"hat"j" sin omega"t"]` ....(1) - Velocity of the particle is given as rate of change of position vector.
∴ `vec"v" = (vec"dr")/"dt" = "d"/"dt" ["r"hat"i" cos omega"t" + "r" hat"j" sin omega"t"]`
`= "r"["d"/"dt" cos omega"t"]hat"i" + "r"["d"/"dt" sin omega"t"]hat"j"`
∴ `vec"v" = - "r"omega hat"i" sin omega"t" + "r"omega hat"j" cos omega"t"`
∴ `vec"v" = "r"omega (-hat"i" sin omega"t" + hat"j" cos omega"t")` - Further, instantaneous linear acceleration of the particle at instant t is given by,
`vec"a" = (vec"dv")/"dt" = "d"/"dt"["r"omega(- hat"i" sin omega "t" + hat"j" cos omega"t")]`
`= "r"omega ["d"/"dt" (- hat"i" sin omega"t" + hat"j" cos omega "t")]`
`= "r"omega ["d"/"dt" (- sin omega"t")hat"i" + "d"/"dt"(cos omega "t")hat"j"]`
`= "r"omega(-omega hat"i" cos omega"t" - omega hat"j" sin omega "t")`
`= - "r"omega^2(hat"i" cos omega "t" + hat"j" sin omega"t")`
∴ `vec"a" = - omega^2("r"hat"i" cos omega"t" + "r" hat"j" sin omega "t")` ....(2) - From equation (1) and (2),
`vec"a" = - omega^2vec"r"` ....(3)
The negative sign shows that the direction of acceleration is opposite to the direction of the position vector. Equation (3) is the centripetal acceleration. - The magnitude of centripetal acceleration is given by,
a = `omega^2"r"` - The force providing this acceleration should also be in the same direction, hence centripetal.
∴ `vec"F" = "m"vec"a" = - "m"omega^2vec"r"`
This is the expression for the centripetal force on a particle undergoing uniform circular motion. - Magnitude of F = `"m"omega^2"r" = "mv"^2/"r"` = mωv
APPEARS IN
RELATED QUESTIONS
A particle rotates in U.C.M. with tangential velocity V along a horizontal circle of diameter ‘D' . Total angular displacement of the particle in time 't' is..........
Draw a neat labelled diagram of conical pendulum. State the expression for its periodic time in terms of length.
A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and direction of acceleration of the stone?
Read the statement below carefully and state, with reason, if it is true or false:
The velocity vector of a particle at a point is always along the tangent to the path of the particle at that point.
Read the statement below carefully and state, with reason, if it is true or false:
The acceleration vector of a particle in uniform circular motion averaged over one cycle is a null vector.
Is it possible to have an accelerated motion with a constant speed? Name such type of motion.
| A small pebble tied at one end of a string is placed near the periphery of a circular disc, at the centre of which the other end of the string is tied to a peg. The disc is rotating about an axis passing through its centre. |
- What will be your observation when you are standing outside the disc? Explain.
- What will be your observation when you are standing at the centre of the disc? Explain.
State True or False
The earth moves around the sun with a uniform.
A uniform metre rule of mass 100g is balanced on a fulcrum at mark 40cm by suspending an unknown mass m at the mark 20cm.
To which side the rule will tilt if the mass m is moved to the mark 10cm ?
Is it possible to have an accelerated motion with a constant speed? Explain.
Solve the following problem.
A projectile is thrown at an angle of 30° to the horizontal. What should be the range of initial velocity (u) so that its range will be between 40m and 50 m? Assume g = 10 m s-2.
What is meant by uniform circular motion? Give two examples of uniform circular motion.
The ratio of angular speed of a hour-hand to the second-hand of a watch is ____________.
A particle is moving in uniform circular motion with speed 'V' and radius 'R'. The angular acceleration of the particle is ______.
A wheel is 0.25 m in radius. When it makes 15 revolutions per minute, its linear speed at the point on circumference is ____________.
A mass 'm' is tied to one end of a spring and whirled in a horizontal circle with constant angular velocity. The elongation in the spring is 1 cm. If the angular speed is doubled, the elongation in the spring is 6 cm. The original length of the spring is ______.
A string of length `l` is fixed at one end and carries a mass 'm' at the other end. The mass is revolving along a horizontal circle of radius 'r' making 'θ' as the semi-vertical angle of cone and `(1/pi)` revolutions per second around the vertical axis through fixed end. The tension in the string is ______.
The given graph represents motion with ______ speed.
Which of the following is correct about uniform circular motion
- the direction of motion is continuously changed
- the direction of motion is not changed
- speed and direction both remain constant
- speed is constant but the direction is changing
A body is said to be in nonuniform motion if it travels ______.
Define uniform circular motion and give an example of it. Why is it called accelerated motion?
A point object moves along an arc of a circle of radius 'R'. Its velocity depends upon the distance covered 'S' as V = `Ksqrt(S)` where 'K' is a constant. If 'e' is the angle between the total acceleration and tangential acceleration, then
A flywheel at rest is to reach an angular velocity of 24 rad/s in 8 second with constant angular acceleration. The total angle turned through during this interval is ______.
For a particle performing uniform circular motion, choose the correct statement(s) from the following:
- Magnitude of particle velocity (speed) remains constant.
- Particle velocity remains directed perpendicular to radius vector.
- Direction of acceleration keeps changing as particle moves.
- Angular momentum is constant in magnitude but direction keeps changing.
A wheel rotating at the same angular speed undergoes constant angular retardation. After the revolution, angular velocity reduces to half its initial value. It will make ______ revolution before stopping.
The distance of the Sun from earth is 1.5 × 1011 m and its angular diameter is (2000) s when observed from the earth. The diameter of the Sun will be ______.
Explain the meaning of uniform circular motion.
By applying Kepler's third law to the formula for centripetal force, Newton concluded that the force acting on a planet is:
A particle moving with uniform speed in a circular path maintains ______.
