For the system shown in figure, `M=1kg` `m=0.2` kg, `r=0.2m` calculate `(g=10m//s^(2))`
(a). The linear acceleration of hoop,
(b). The angular acceleration of the hoop of mass `M` and
(c). The tension in the rope.

In the figure ( g = 10 m//s^(2) ) .Acceleration of 2kg block is :

In the figure shown, the acceleration of block of mass m_(2) is 2m//s^(2) . The acceleration of m_(1) :

In the figure shown, the acceleration of block of mass m_(2) is 2m//s^(2) . The acceleration of m_(1) :

A 20kg monkey slides down a vertical rope with a constant acceleration of 7m//s^2 . If g =10 ms^(-2) , What is the tension in the rope ?

A 20 kg monkey slides down a vertical rope with a constant acceleration of 7 m/ s^2 . If g = 10 m s^-2 . What is the tension in the rope ?

A 20 kg monkey slides down a vertical rope with a constant acceleration of 7 ms^(-2) . If g = 10 m//s^(-2) , what is the tension in the rope ?

In the system shown in figure m_(B)=4kg , and m_(A)=2kg . The pulleys are massless and friction is absent everywhere. The acceleration of block A is.

In the figure shown ,find : (g = 10 m//s^(2)) (a) acceleration of 1 kg,2 kg and 3 kg blocks and (b) tension T_(1) and T_(2) .

In the figure shown ,find : (g = 10 m//s^(2)) (a) acceleration of 1 kg,2 kg and 3 kg blocks and (b) tension T_(1) and T_(2) .

Two blocks shown in figure are connected by a heavy uniform rope of mass 4 kg . An upward force of 200 N is applied as shown (a) What is the acceleration of the system . (b)What is the tension at the top of the rope ? ( c) What is the tension at the mid- point of the rop (Take g = 9.8 m//s^(2))