1) The velocity of cylinder due to left wedge is $-v/4 \hat i - \sqrt3 v/4 \hat j$

Velocity of cylinder due to right wedge is $2u \hat i$

So net velocity is $\sqrt{13}/2$

2) Can we also do it by assuming right wedge at rest.

1 vote

1) The velocity of cylinder due to left wedge is $-v/4 \hat i - \sqrt3 v/4 \hat j$

Velocity of cylinder due to right wedge is $2u \hat i$

So net velocity is $\sqrt{13}/2$

2) Can we also do it by assuming right wedge at rest.

Yes you can use any frame of reference which you find convenient. There is no acceleration so a constant velocity frame is appropriate. You can use a frame at rest is either wedge or in the cylinder.

Sorry I don't know. The method used in the given solution looks good to me. If you want to use a more difficult method, that is your choice, but it does not seem sensible to me. Unless you have a difficulty with the given solution, I shall leave the question "what is my mistake?" to somebody else.

Using a frame of reference in which one of the wedges is at rest is probably the easiest solution. If the left wedge is at rest, the motion of the cylinder is just sliding/rolling down a fixed inclined plane. If the right wedge is at rest, the cylinder is sliding/falling vertically (but not in free fall). Using any other frame of reference makes the motion of the cylinder more difficult to visualise.

Using a frame of reference in which one of the wedges is at rest is probably the easiest solution. If the left wedge is at rest, the motion of the cylinder is just sliding/rolling down a fixed inclined plane. If the right wedge is at rest, the cylinder is sliding/falling vertically (but not in free fall). Using any other frame of reference makes the motion of the cylinder more difficult to visualise.

1 vote

Best answer

You can use any frame of reference which you want. But it is best to look for a frame which makes the problem easier to solve.

Using a frame of reference fixed in the right-hand wedge is no more difficult than one fixed in the left-hand wedge. In this frame the left wedge moves left with speed $3u$ and the cylinder moves vertically downward with speed $v$.

In time $t$ the wedge moves left by distance $x=3ut$. The point of contact with the cylinder moves down by distance $y$ such that

$\tan30^{\circ}=\frac{1}{\sqrt3}=\frac{y}{x}$

so

$y=\frac{x}{\sqrt3}=\frac{3ut}{\sqrt3}=\sqrt3 ut$.

The vertical speed of the cylinder in this frame is

$v=\frac{y}{t}=\sqrt3u$.

Transforming back into the laboratory frame, the cylinder has velocity $2u$ horizontally to the right and $\sqrt3u$ vertically down. Using Pythagoras' Theorem the resultant velocity is

$\sqrt{4u^2+3u^2}=\sqrt7u$.

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