Question

A spring which obey’s Hooke’s law extends by 1 cm when a mass is hung on it. It extends by a further 3 cm when the attached mass is moved in a horizontal circle making 2 revolutions per second. What is the length of the unstretched spring? Take 

Solution

Correct option is

21 cm

 

According to Hookes’ law, the stretching force F = kx, where k is the force constant and x, the extension of the spring. A force mg stretches the spring by 1 cm. When the mass is describing the horizontal circle, total stretching = 1 + 3 = 4 cm. Hence

                                    

Referring to fig, the horizontal component T sin θ provides the necessary centripetal force for circular motion, i.e.  

                   

. Let L cm be the length of the unstretched spring. Then AC = (L + 4) cm and r = (L + 4) sin θ.     

  

                    

  

or L = 25 – 4 = 21 cm.

SIMILAR QUESTIONS

Q1

 

A simple pendulum of length L and with a bob of mass M is oscillating in a plane about a vertical line between angular limits

–Ï• and +Ï•. For an angular displacement θ (< Ï•) the tension is the string and the velocity of the bob are T and v respectively. The flowing relations hold good under the above conditions:

Q2

A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle. The motion of the particle takes place in a plane. It follows that:

Q3

A simple pendulum of bob mass m swings with an angular amplitude of 40o. When its angular displacement is 20o, the tension in the string is

Q4

A car moves at a speed of 36 km h–1 on a level road. The coefficient of friction between the tyres and the road is 0.8. The car negotiates a curve of radius R. If g = 10 ms –2, the car will skid (or slip) while negotiating the curve if value of R is  

Q5

 

A train has to negotiate a curve of radius 200 m. by how much should the outer rails be raised with respect to the inner rails for a speed of 36 km h –1. The distance between the rails is 1.5 m.

Take g = 10 ms–2

Q6

One end of a string of length R is tied to a stone of mass m and the other end to a small pivot on a frictionless vertical board. The stone is whirled in a vertical circle with the pivot as the centre. The minimum speed the stone must have, when it is at the topmost point on the circle, so that the string does not slack is given by

Q7

The pilot of an aircraft, who is not tied to his seat, can loop a vertical circle in air without falling out at the top of the loop. What is the minimum speed required so that he can successfully negotiate a loop or radius 4 km? Take g = 10 ms–2 

Q8

One end of a string of length 1.0 m is tied to a body of mass 0.5 kg. It is whirled in a vertical circle as shown in fig. If the angular frequency of the body is 4 rad s–1, what is the tension in the strong when the body is at the topmost point A? Take g = 10 ms–2.

                                                                       

Q9

A simple pendulum of length r and bob mass m swings in a vertical circle with angular frequency ω. When the string makes an angle θ with the vertical, the speed of the bob is v. The radical acceleration of the bob at this instant is given by

Q10

A boy whirls a stone in a horizontal circle 2 m above the ground by means of a string 1.25 m long. The string breaks and the stone flies off horizontally, striking the ground 10 m away. What is the magnitude of the centripetal acceleration during circular motion? Take g = 10