Recent content by ScienceIsMyLady

Thank you. I looked it up and I think you are right. "impulse" is the right term. I learned this stuff ages ago so I made mistake on their names...

Thanks for your remark

Thanks a lot.

Would you mind telling me other than mg cos \theta, what force has been exerted on the dome by the particle?

Thanks for your answer. What I said "due to gravity" is the situation you described here. The particle exerts force on the dome because it has got pulled by the gravitational force by the Earth and the dome is stopping it. I didn't mean "the gravitational force exerted by the Earth on the...

Sorry, I didn't say clearly. I meant impact on the particle, because there is change in velocity, hence momentum. This force should be exerted by the dome. In return, the particle also exerts an equal but opposite force on the dome, as stated in Newton's third law. Is that right? Thanks.

Did you mean that when the velocity of the particle is changed, an impact force is exerted on the particle by the dome and an equal but opposite force is exerted on the dome by the particle, so the normal reaction R is not just the gravitational force component but also includes this impact...

I agree with you that the surface is preventing the particle from falling vertically through the surface and this force together with gravity forms the trajectory of the particle. But I don't understand why the resulting net force isn't tangential. While it is proven to be true that fixed...

Thank you for your answer. But I still don't quite understand why it is necessary, because if the particle does not fit the condition for circular motion, then it is not in a circular motion. In fact, it does fall off eventually. Nothing is really keeping it in a circular path.

A particle was resting on top of a dome and given a negligible push such that it falls. The question is at what angle will the particle fall off the dome. The solution is that, m\frac{v^2}{a} = mg\cos\theta - R and by conservation of energy 0 + mga = \frac{1}{2}mv^2 + mga\cos\theta then the...

Perhaps it means with the same rotational kinetic energy? \frac{1}{2}I\omega^2 With the same rotational kinetic energy, if you want to have higher angular velocity, you want a lower moment of inertia, which does depend on the mass distribution