Cool Pulley

Classical Mechanics Level 4

The figure above shows a block B of negligible mass connected to a particle of mass $1 \text{ kg}$ with a non-stretchable string. The hanging particle, at rest, is given a velocity $\sqrt{8gH} \text{ m/s}$ horizontally. The initial acceleration of block B is $K \text{ m/s}^2$ . Find the value of $K$ .

Details:

Acceleration due to gravity $g=10 \text{ m/s}^2$ .
$H=2 \text{ m}$ .

The answer is 50.

1 solution

Brilliant Physics Staff
Jun 21, 2016

The first thing to realize is that the block is massless, so it will have the same acceleration as the mass (since they're connected by the inextensible string).

When the mass is first pushed out to the side, it is pulled down by gravity, but it also begins to swing, giving rise to a centripetal acceleration toward the vertical.

Therefore, the total acceleration of the mass-string-block system is given by

$\begin{aligned} a &= g + v^2/R \\ &= g + \frac{\left(\sqrt{8gH}\right)^2}{2H} \\ &= g + 4g \\ &= 5g \end{aligned}$

Can we use constraint relations?

Priyanshu Mishra - 4 years, 11 months ago

What do you mean?

Brilliant Physics Staff - 4 years, 11 months ago

I mean that can i fix something like length of string in terms of given quantities and then differentiate or integrate to get the result?

Also, in your solution ; you have written

When the mass is first pushed out to the side, it is pulled down by gravity, but it also begins to swing, giving rise to a centripetal acceleration toward the vertical.

SIR, can you explain me why the mass begins to swing and rotate giving rise to centripetal acceleration?

Priyanshu Mishra - 4 years, 11 months ago

@Priyanshu Mishra – Initially the y component of velocity = 0 and x component of velocity is present . We know that a body moves in the direction of velocity and not the direction of force therefore the body initially moves only in x direction . At any instant the acceleration component which is perpendicular to velocity is the normal acceleration which is solely responsible for the necessary centripetal acceleration to change the direction of velocity . Hence the resultant of tension and weight (forces perpendicular to velocity) is the provide the normal acceleration = $\dfrac{(V tangential)^2}{radius}$ . At the given instant radius = 2H . This explains why it initially starts rotating .

space sizzlers - 4 years, 9 months ago

But isn't the centripetal acceleration in the opposite direction of the gravitational acceleration?

Anupam Nayak - 4 years, 11 months ago

The force on the mass is upward, but the force on the block is in the opposite direction.

Josh Silverman Staff - 4 years, 6 months ago

What do you mean by this? Pls explain it asap.

Om Tiwari - 2 years, 10 months ago

@Om Tiwari – I see that's a little hard to parse. What I mean is

the centripetal acceleration points upward.
but that means that the force that the rope feels from the hanging mass is downward.
and then (following the direction of the force around the rope) the force on the block B is toward the right.

Let me know if that's still unclear.

Josh Silverman Staff - 2 years, 10 months ago

@Josh Silverman – Help with this https://brilliant.org/discussions/thread/vector-analysis/

Rishi Tiwari - 2 years, 9 months ago

Cool Pulley

The answer is 50.

1 solution

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