Recent content by AakashR

Impulse, I=F*t= change in momentum= mΔv. I guess now you know how to solve the problem. Cheers

Ramanujan, Gauss, Newton, Riemann, Einstein

you need more information to solve the problem. you need to know the tension in the rod or string that's supporting the pendulum. Then you will be able to solve the eqn ΣF=ma=mg-T Cos θ

v(t)= -(0.347 m/s)sin(15.0t + 2.00π) is equal to v(t)= -(0.347 m/s)sin(15.0t). Now attempt to solve it in the same way as before and you will get t=0.0195s which is the correct answer.

Remember one thing- All the oscillations have two things in common I. the oscillation takes place about an equilibrium position and, II. the motion is periodic. These are the only two conditions that are to be fulfilled for a motion to be oscillatory motion.

If the total distance covered by the weight while bobbing up and down is 10cm then magnitude of the amplitude is half the total distance that is 5cm.

Here is the answer- We can see that both the weight of the block and tension in the rope are constant. So, acceleration of the block is going to remain constant. let's consider vertically downward direction as +Y. We have ΔY=1.5m and Δt=2.0s. Now, by using the formula ΔY=0.5a(Δt)^2, we...

dude the inner radius is 0.06m and outer is 0.08m. you should subtract the value of MIdisk you get for 0.08m from the one you get for 0.06m. Why did you add the two up. It's not making any sense. After you do it you can solve the problem easily by using conservation of energy.