Question

Please help with this physics problem?

A small ball of mass M is connected to a block of mass 2M by a light extensible string of length R.The block and the ball are kept on the round.There is sufficient friction to prevent the block from slipping.The ball is projected vertically up with a velocity U.The velocity U for which the centre of mass of the block and the ball moves in a circle satisfies the equation:
`(a)3gr<=U^2>=2gr`
`(b)U^2>=5gr`
`(c)2gr<=U^2<=5gr`
`(d)"None " "of" " these"`
please explain the solution in detail.

Answer 1

None of the above.
The correct condition is
`3gR <= U^2 <= sqrt{8/3}3gR`

Explanation:

I am assuming that the initial configuration was one in which the string is stretched out straight so that the ball is at distance `R` from the block (this is the only case where the ball can move in a circle after being projected vertically.)

I am also assuming that the words "moves in a circle" actually means "moves in a complete semicircle". After all, due to the presence of the ground, it can not describe anything more than a semicircle . On the other hand "complete" is necessary - otherwise no matter what the velocity, the center of mass will, at least initially, describe a part of a circle.

The center of mass will move along a complete semicircle if two conditions are satisfied

1. the ball moves along a complete semicircle.
2. the block stays fixed on the ground throughout
   (under these conditions the center of mass will move all the way along a semicircle of radius `R/3` )

For this, the string must be taut throughout (the tension in the string must at least be zero), and the upward component of the tension can never be more than `2mg`.

If the velocity is `v` when the string makes an angle of `theta`
with the vertical, we have :

From conservation of energy :
`1/2 mv^2+mgR cos theta = 1/2mU^2 implies v^2 = U^2-2gR cos theta`

From Newton's second law (for the component along the string)

`T+mg cos theta = {mv^2}/R = {mU^2}/R-2mg cos theta implies`
`T = {mU^2}/R-3mgcos theta`

The minimum value that `T` reaches is thus `{mU^2}/R-3mg`, and since this can not be negative (a string can only pull - it can't push).

So `U^2 \ge 3gR`

On the other hand, the vertical component of the tension at `theta` is given by

`T cos theta = ( {mU^2}/R-3mgcos theta) cos theta = 3mg({U^2}/{3gR}-cos theta) cos theta`

We need to ensure that the largest value of this component is not more than `2mg`.

It is easy to maximize an expression of the form `(a-x)x` by elementary algebra

`(a-x)x = -(x^2-2 a/2 x+a^2/4)+a^2/4 = a^2/4 - (x-a/2)^2`

so that the maximum value is `a^2/4`, attained at `x = a/2`

So, the largest value of `T cos theta` is `3mg times 1/4 ({U^2}/{3gR})^2`, and since this can not exceed `2mg`, we have

`({U^2}/{3gR})^2 <= 8/3`

and so `U^2 <= sqrt{8/3}3gR`

Thus the condition that one is aiming for is

`3gR <= U^2 <= sqrt{8/3}3gR`

Note : it may be mistakenly assumed that `T cos theta` is largest where `cos theta` is maximum, namely, at the top most point. This would have lead to the condition

`{mv^2(0)}/R = T+mg <= 2mg+mg = 3mg`

and this would have led to `v^2(0) <= 3mg` which would lead to

`U^2 <= 5gr`

Answer 2

None of these.

Explanation:

Assuming that the initial string elongation is `r` the initial velocity `U` in the vertical and that the angle with the horizontal axis is `theta` we have

`1/2mU^2=1/2m v^2+rm g sin theta`

In the attached picture can be depicted the main elements involved. In red

`m v^2/r` centrifugal force and `m r ddot theta` accelerated mass along the circular path.

the tension `T` is given as

`T = m v^2/r - m g sin theta`

after substitution

`T = m(U^2/r - 3g sin theta)`

then `T > 0 rArr U^2/r - 3g sin theta > 0` or

`sin theta < U^2/(3rg)`

then if we choose `U` such that `U^2/(3rg) > 1` the tension `T` will be maintained all along the path.

Considering the normal reaction in the base body `N`

`N = lambda m g - T sin theta = m(lambda g - (U^2/r - 3g sin theta)sin theta)`

Here `lambda = 2`. Now if we want a circular path we need `N > 0` or

`lambda g - (U^2/r - 3g sin theta)sin theta > 0` or

`sin theta = (U^2 pm sqrt[U^4-12 g^2 lambda r^2 ])/(6 g r)`

and then

`sin theta < (U^2 - sqrt[U^4-12 g^2 lambda r^2 ])/(6 g r)` and
`sin theta > (U^2 + sqrt[U^4-12 g^2 lambda r^2 ])/(6 g r)`

Considering now that `sin theta le 1` if we choose `U` such that

`(U^2 - sqrt[U^4-12 g^2 lambda r^2 ])/(6 g r) > 1` then then along the path we will have `N > 0` as desired, or

`U^2 < (3+lambda)r g`

Resuming, the condition for a neat path is

`3 r g < U^2 < (3+lambda) r g`