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CIRCULAR MOTION Sol. If v is instantaneous velocity, centripetal accel erati on a c = 2 v r ⇒ 2 v r = k 2 rt 2 ⇒ v = krt In circular motion work done by centripetally force is always zero & work is done only by tangential force. Q Tangent acceleration a t = dv d dt dt = (krt) = kr ∴ Tangential force F t = ma t = mkr Power P = F t v = (mkr) (krt) = mk 2 r 2 t 3 . CENTRIPETAL AND CENTRIFUGAL FORCE 3.1 Centri petal force : In uniform circular motion the force act ing on the particle along the radius and towards the centre keeps the body moving along the circular path . This force is called centripetal force. Explanation : (i) Centripetal force in necessary for uniform circular motion. (ii) It is along the radius and towards the centre. (iii) Centripetal force = [mass] × [centripetal acceleration] = 2 mv r = mr ω 2 (iv) Centripetal force is due to known interaction. Therefore it is a real force. õ If an object tied to a string its revolved uniformly in a horizontal circle, the centripetal force is due to the tension imparted to the string by the hand. õ When a satellite is revolving in circ ular orbit round the earth, the centripetal force is due to the gravitational force of attraction between the satellite and the earth. õ In an atom, an electron revolves in a circular orbit round the nucleus. The centripetal force is due to the electrostatic force of attraction between the positively charged nucleus and negatively charged EXAMPLE BA EXAMPLE BA EXAMPLE BA EXAMPLE BA EXAMPLE BA SED ON CENTRIPET SED ON CENTRIPET SED ON CENTRIPET SED ON CENTRIPET SED ON CENTRIPET AL FO RC E AL F OR CE AL FO RC E AL F OR CE AL FO R CE Ex.7 Stone of mass 1kg is whirled in a circular path of radius 1 m. Find out the tension in the string if the linear velocity is 10 m/s ? Sol. Tension 2 mv R = 2 1 ( 10) 1 × = 100 N Ex.8 A satellite of mass 10 7 kg is revolving around the earth with a time period of 30 days at a height of 1600 km. Find out the force of attraction on satellite by earth? Sol. Force = mω 2 R and 2 2 3.4 T 30 86400 π × = × = 6 6.28 2.59 10 × Force = mω 2 r = 2 6 6.28 2.59 10 × × 10 7 × (6400 + 1600) × 10 3 = 2.34 × 10 6 N v v v
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CIRCULAR MOTION
Sol. If v is instantaneous velocity, centripetal acceleration ac =
2v
r
2v
r = k2 rt2 v = krt
In circular motion work done by centripetally force is always zero & work is done only by tangential force.
Q Tangent acceleration at = dv ddt dt
= (krt) = kr
Tangential force Ft = mat = mkr
Power P = Ftv = (mkr) (krt) = mk2r2t3 . CENTRIPETAL AND CENTRIFUGAL FORCE
3.1 Centripetal force : In uniform circular motion the force acting on the particle along the radius and towardsthe centre keeps the body moving along the circular path . This force is called centripetal force.
Explanation :
(i) Centripetal force in necessary for uniform circular motion.(ii) It is along the radius and towards the centre.
(iii) Centripetal force = [mass] [centripetal acceleration] = 2mv
r = mr2
(iv) Centripetal force is due to known interaction. Therefore it is a real force.
If an object tied to a string its revolved uniformly in a horizontal circle, the centripetalforce is due to the tension imparted to the string by the hand.
When a satellite is revolving in circular orbit round the earth, the centripetal forceis due to the gravitational force of attraction between the satellite and the earth.
In an atom, an electron revolves in a circular orbit round the nucleus. The centripetal force is due to theelectrostatic force of attraction between the positively charged nucleus and negatively charged
EXAMPLE BAEXAMPLE BAEXAMPLE BAEXAMPLE BAEXAMPLE BASED ON CENTRIPETSED ON CENTRIPETSED ON CENTRIPETSED ON CENTRIPETSED ON CENTRIPETAL FORCEAL FORCEAL FORCEAL FORCEAL FORCE
Ex.7 Stone of mass 1kg is whirled in a circular path of radius 1 m. Find out the tension in the string if thelinear velocity is 10 m/s ?
Sol. Tension 2mv
R =
21 (10)1
= 100 N
Ex.8 A satellite of mass 107 kg is revolving around the earth with a time period of 30 days at a height of 1600km. Find out the force of attraction on satellite by earth?
Sol. Force = m2R and 2 2 3.4T 30 86400pi
=
= 6
6.282.59 10
Force = m2r = 2
66.28
2.59 10
107 (6400 + 1600) 103 = 2.34 106 N
v
v
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