A simple pendulum having a bob of mass m is suspended from the ceiling of a car used in a stunt film shooting. The car moves up along an inclined cliff at a speed v and makes a jump to leave the cliff and lands at some distance. Let R be the maximum height of the car from the top of the cliff. The tension in the string when the car is in air is
A simple pendulum having a bob of mass m is suspended from the ceiling of a car used in a stunt film shooting. The car moves up along an inclined cliff at a speed v and makes a jump to leave the cliff and lands at some distance. Let R be the maximum height of the car from the top of the cliff. The tension in the string when the car is in air is
mg - 
mg + 
The correct option is D zero
When a car goes off cliff its like a projectile being throw with an initial velocity of u m/s and some angle of projection θ.
Now we know such a projectile will always be under only one acceleration i.e. 'g' downwards. So all the parts in the car (the roof, the pendulum) have the same velocity and same acceleration downwards i.e. 'g'
You should be able to say intuitively that since both the car's ceiling and pendulum are under free fall there can't be any tension in the string as it won't be taut.
Even if you are not then draw free body diagram of bob assuming some tension T
⇒mg−T=ma
a=g
⇒mg−T=mg
⇒T=0
zero
When a car goes off cliff its like a projectile being throw with an initial velocity of u m/s and some angle of projection θ.
Now we know such a projectile will always be under only one acceleration i.e. 'g' downwards. So all the parts in the car (the roof, the pendulum) have the same velocity and same acceleration downwards i.e. 'g'
You should be able to say intuitively that since both the car's ceiling and pendulum are under free fall there can't be any tension in the string as it won't be taut.
Even if you are not then draw free body diagram of bob assuming some tension T
⇒mg−T=ma
a=g
⇒mg−T=mg
⇒T=0
