A uniform disc of mass m, radius R is placed on a smooth horizontal surface. If we apply a horizontal force F at P as shown in the figure. If F = 4 N, m= .1 kg, R = 1 m and r = 1/2 m then, find the: A rod of length l = 1 m leaning against a vertical wall is pulled at its lowest point A with a constant velocity v = 4 ms^(-1). In consequence, the rod rotates win the vertical plane. When the rod makes an angle theta=4ms^(-1) with vertical and find the angular velocity of the rod (in rad s^(-1)).

A uniform disc of mass m, radius R is placed on a smooth horizontal su

1.	A uniform disc of mass m, radius R is placed on a smooth horizontal surface. If we apply a horizontal force F at P as shown in the figure. If F = 4 N, m= .1 kg, R = 1 m and r = 1/2 m then, find the: A rod of length l = 1 m leaning against a vertical wall is pulled at its lowest point A with a constant velocity v = 4 ms^(-1). In consequence, the rod rotates win the vertical plane. When the rod makes an angle theta=4ms^(-1) with vertical and find the angular velocity of the rod (in rad s^(-1)).
Answer» Solution :From the geometry of the arranement we observe that the ends `A` and `B` side on the horizontal and VERTICAL surfaces with velocities `v_(A)` and `v_(B)` respectively, where `v_(A)=v` (given) and `v_(B)` is assumed. As per the procedure, let US drop perpediculars `AX` and `BY` from `A` and `B` respectively onto their lines of motion (`vecv_(A)` and `vecv_(B)`). We observe that the perpendicular intersect at `P`. Then `P` is the instantaneous AXIS of ROTATION. Since `v_(P)=0` we have `omega_(BP)=omega_(AP)(=omega)` substituting `PA=lcostheta` and `v_(A)=v` We have `omega=v/(costheta)=4/(1.4/5)=5rads^(-1)`