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1

It is not necessary to consider the torque on the ball. It only makes the situation more difficult to think about. It is best to stick to energy considerations. Because the ball does not slip, the friction force does no work. However, contrary to your assumption, this does not mean that rotational KE is constant. This is where your reasoning is going ...

3

$$F_1=mg\cos\theta$$ $$F_3=mg\sin\theta$$ With no radial notion and Newton's 2nd: $$F_2=F_1$$ Using the simple model of friction, with $\mu$ the static coefficient of friction (no slipping is assumed): $$F_f=\mu F_2=\mu mg \cos\theta$$ Call $R$ is the radius of the sphere and $r$ the radius of the ball. The torque $\tau_1$ caused by $F_3$ about $O$ ...

1

There are mathematical ways of showing all this, but it is important to have intuition before starting to calculate. Here is my intuitive picture: The kinetic energy of the rolling cylinder has two components: The kinetic energy of its motion along the plane. The kinetic energy of its rotation about its own axis. The only source of kinetic energy is the ...

0

Thank you for the update, WJ47. The slope of the blue tube looks very steep. Both the ruler/tube and the white cylinder (cellotape holder) look quite smooth, so I think there will be little friction, resulting in a mixture of rolling and sliding here. It is very difficult to predict how much of each. This is a rather 'messy' experiment, IMHO, difficult ...

2

It is true that the weight will pull the machine to the north when it points north because the machine is pulling the weight to the south; by Newton's Third Law, that means the weight is pulling the machine northward. However, you have to think about what happens during the rest of the rotation. To have a faster velocity while the weight points north, the ...

6

Here is a video that shows the ball behaviour you're describing. The phenomenon is explained by the CoandÄƒ effect: the tendency of a fluid jet to stay attached to a convex surface. Note that the ball does actually move in a kind of oscillatory motion. This is probably due to the water jet in the video not being highly stable. When more liquid is running ...

2

Take a rope with a small weight on one end and spin around on a point such that the rope is taut. You will feel force on your arm 'pulling' on it towards the weight. That force exists because that's what you are exerting on the object to prevent constantly change it's direction. The value you refer to is the acceleration portion of that force from F=ma. ...

5

what does this value mean? This value means for maintaining of moon on the circular path, we need a force by amount of $\textrm{mass of the moon}\times\textrm{that value}$ How can this value be interpreted intuitively? According to first law of Newton, a moving body will move on a straight line uniformly unless a force is exerted on it. As you can ...

2

Acceleration is defined as the time rate of change of velocity - be it in its direction or magnitude or both. Velocity $\mathbf v$ in polar coordinates is given by: $$\mathbf v= \dot {\mathbf r} = \dot r\mathbf{e_r} + r~\dot \theta\mathbf{e_{\theta}}\;.\tag 1$$ And then acceleration $\bf a$: \mathbf a = \dot{ \mathbf v} = \left[\ddot r - r\dot \theta ^...

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