Answer to Question #238910 in Mechanics | Relativity for ddd

Question

A horizontal frictionless table has a small hole in its center. Block A on the table is connected to block B hanging beneath by a string of negligible mass which passes through the hole. Initially, B is held stationary and A rotates at constant radius 𝑟𝑟0 with steady angular velocity 𝜔𝜔0. (a) (2 marks) Draw a free body diagram of all the mobile bodies.

(b) (3 marks) Find the radial and tangential component of the acceleration of block A.

(c) (5 marks) Find the acceleration of block B immediately after the release. Also find the tension in the string.

Answer 1

Explanations & Calculations

• Refer to the figure attached for the free body diagram & the rest.

b)

• Since block B is not released yet, block A continues on that circular path.
• As block A has a steady angular velocity, it has no change in tangential velocity thus its tangential acceleration is zero and remains at that as long as the block has the same steady velocity.
• And as we already know (not going to do the proof here) during a circular motion the object is subjected to centripetal acceleration of magnitude $\small a= r\omega^2=\large\frac{v^2}{r}=\small v\omega$ nad for this case it is $\small r_0\omega^2$.

c)

• Immediately after the release, the motion of the block is just perpendicular to the thread thus the motion of block A is perpendicular to the tension/force, therefore, the motion of block A due to B is independent of A's circular motion. Just using the F=ma for the linear motion(motion of A due to B's downward pull) of both blocks we could get the B's acceleration.

$\qquad \qquad \begin{aligned} \small \to F&=\small ma\\ \small T&=\small ma\cdots(1)\\ \small \downarrow F&=\small Ma\\ \small Mg-T&=\small Ma\cdots(2)\\ \\ \small a&=\small \frac{Mg}{(M+m)} \end{aligned}$

• Then the tension would be,

$\qquad \qquad \begin{aligned} \small T&=\small ma\\ &=\small \frac{Mmg}{(M+m)} \end{aligned}$