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.

2 Answers
Feb 19, 2018

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#

Feb 19, 2018

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#

enter image source here

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#