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3

Assuming the wheel moves at the same speed as the water (ie: neglecting and 'slip' of the wheel past the water), angular velocity is given by: $$\omega = v / r$$ where $r$ is the 'radial' distance from the centre of the wheel to the top of the water. In practice, there will be some water sliding past the wheel, depending upon the hydrodynamics of the ...

8

The reason that you get slip at even the smallest forces results not from the fact that the tire is slipping against the ground, but that the tire is elastic. There is no way to completely eliminate slip with an elastic tire. Let's see why this is. To measure the slip, lets put twenty little green splotches of die evenly spaced on the circumference of the ...

0

Both graphs have 0% force at 0% slip. That implies that no force is being applied to the tire. As applied force is increased until a certain point the increase in slip remains nearly linear because the coefficient of friction remains nearly constant. Thus effective force also increases nearly linearly. After that point the coefficient of friction gets ...

2

The tension in the rope should be different on the left than on the right - it is this difference that gives rise to the torque that accelerates the pulley. You seem to think that it should be the same: but if it was, then where would the torque to move the pulley come from? Annotate your diagram carefully: you did not show $T$ anywhere.

2

Your mistake is that the two tensions are different, because of the presence of a pulley with non-zero $I$. What you have missed is: 1) connect the two tensions to the torque: $(T_1-T_2)R=I\alpha$ and 2) link the accelerations $R\alpha=a_1=a_2$ NOTE (from comments): If the pulley had a zero $I$ (moment of inertia), then the two tensions would be be ...

0

I don't know how this is supposed to simulate planetary motion, but we can still look at what data you would expect. The force on the stopper is $\frac {mv^2}r$, which if there is no friction (good luck!) equals the gravitational force on the masses at the bottom. We would therefore expect that for given mass on the bottom, $r \propto v^2$, which doesn't ...

0

Assume that the origin of frame $r$ is moving with acceleration $\newcommand{\a}{\mathbf{a}}\a$, with respect to a fixed frame $f$, but that the coordinate axes of $r$ are aligned with $f$'s. Then if an object has an apparent acceleration $\a_r$ in the frame $r$, then it's actual acceleration $\a_f$ in the fixed frame is $\a_r + \a$. Now when your allow the ...

0

An object with corners or longer in one dimension than in the others, can fall on such a part and be in unstable equilibrium. E.g. an object touches the ground on one of its corners but the center is not on the vertical that passes through that corner. That generates a torque, and angular momentum appears that tends to be conserved.

2

If two eggs are given out of which one is hard boiled egg and the other is raw egg. When both of them are allowed to spin on a tabletop, the egg which spins slower must be the raw egg because the liquid inside it tries to get away from the axis of rotation increasing the value of moment of inertia (I). To remain the angular momentum L conserved in absence of ...

1

what does the commentator mean by concavity of the floor? He or she means that the surface you walk on is in fact the inside of a cylindrical surface. Like a very large version of the inside of a wedding-ring. The curvature of this surface can be measured.

1

That the floor is not flat but circular, an the centrifugal force acts radially, and is the same at the same radius. If you cover the concavity of the floor with a planar surface or either make the stations of planar segments (a polygon instead of a circle), then you will measure different accelerations at different points of the floor, because they are not ...

-1

I have yet to find a physics book that doesn't make this really confusing. If one has a vector fixed in inertial space, its components as viewed in a moving frame are obtained by the dot product of the vector with the moving unit triad fixed to the body but moving relative to inertial space. While the inertial frame would measure its components as constants ...

0

The key phrase to notice here is that you are being asked for the least horizontal force. If you use a different expression for torque, you should be able to find the least horizontal force by figuring out what the minimum angle needed is. Remember: $$\tau=\underline{\mathbf{r}}\times\underline{\mathbf{F}}$$

-1

If we look at the symmetry of the pear of mass $m$, we can see on the x axis that momentum is conserved because the derivative of its motion in space $x$ is the same from any point we measure it's motion from: $$p_x=m \frac{dx}{dt}$$ On the y axis however, the system is affected by an external field gravity, $g$ which pulls the pear downwards. This removes ...

-1

There are two types of vector.One is polar,the other is axial.Angular velocity is an axial vector.So,no displacement is needed along its direction.

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