Categories

## Maximum Height of Object over a Pulley

Try this problem, useful for Physics Olympiad based on Maximum Height of Object over a Pulley.

The Problem: Maximum Height of Object over a Pulley

Two objects with masses 5Kg and 2Kg hang 0.6m above the floor from the ends of a cord 6m long passing over a frictionless pulley. Both objects start from rest. Find the maximum height reached by the 2.00Kg object.

Discussion:

Set up : After the (5Kg) object reaches the floor, the (2Kg) object is in free fall, with downward acceleration (g).

Execution: The (2Kg) will accelerate upward at$$\frac{5-2}{5+2}g=3g/7$$ and the (5Kg) object will accelerate downward at (3g/7).

Let the initial height above the ground be (h_0).

When the large object hits the ground, the small object will be at a height (2h_0) and moving upward with a speed given by $$v_0^2=2ah_0=6gh_0/7$$. The small object will rise to a distance (v_0^2/2g=3h_0/g) and so the maximum height reached will be $$2h_0+3h_0/7=17h_0/7=1.46m$$ above the floor, which is (0.860m) above its initial height.

Categories

## A Pulley System

Let’s discuss a problem, useful for Physics Olympiad based on A Pulley System. Read the problem carefully, find it yourself, and then read the solution.

The Problem: A Pulley System

One end of a string is attached to a rigid wall at point O, passes over a smooth pulley, and carries a hanger S of mass M at its other end.  One end of a string is attached to a rigid wall at point O, passes over a smooth pulley, and carries a hanger S of mass M at its other end. Another object P of mass M is suspended from a light ring that can slide without friction, along the string, as is shown in the figure. OA is horizontal. Find the additional mass to be attached to the hanger S so as to raise the object P by 10cm.

Solution:

Let us denote the tension in each string as T. $$2Tcos\theta=Mg$$$$2(Mg)cos\theta=Mg$$$$cos\theta=\frac{1}{2}$$$$\theta=60^\circ$$$$tan60=\frac{\frac{40\sqrt{3}}{2}}{PQ}$$$$tan60^\circ=\sqrt{3}$$Hence,$$PQ=20cm$$Now, when an additional mass m is hung from the pulley, the length of PQ changes to P’Q’.

$P’Q’=PQ-10=20-10=10$.

$$Q’S’=\sqrt{P’Q’^2+P’S^2}=\sqrt{1300}$$Now, again considering the force equation$$2Tcos\theta=Mg$$$$2(M+m)g\times\frac{10}{\sqrt{1300}}=Mg$$$$2(M+m)\times\frac{1}{\sqrt{13}}=M$$$$2(M+m)=\sqrt{13}M$$$$2m=M(\sqrt{13}-2)$$$$m=\frac{M\times(\sqrt{13}-2)}{2}=0.9M$$

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## Mass supported by a Hollow Cylinder

Let’s discuss a beautiful problem useful for Physics Olympiad based on Mass supported by a Hollow Cylinder.

The Problem:

A mass m is supported by a massless string wound on a uniform hollow cylinder of mass m and radius R. If the string does not slip on the cylinder, with what acceleration will the mass fall on release?

Solution:

For the mass m, the force equation stands as:

$$mg-T=ma….(i)$$

For the cylinder, the force equation is:

$$T.R=mR^2(a/R)$$

Hence,

from above equation $$T=ma$$

Now, putting the value of T in equation (i), we get

$$ma=2mg$$

$$\Rightarrow a=g/2$$

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## Masses over a frictionless pulley

Let’s discuss a beautiful problem useful for Physics Olympiad based on masses over a frictionless pulley.

The Problem: Masses over a frictionless pulley

Two bodies A and B hanging in the air are tied to the ends of a string which passes over a frictionless pulley. The masses of the string and the pulley are negligible and the masses of two bodies are 2kg and 3kg respectively. (Assume g=(10m/s^2)). Body A moves upwards under a force equal to
(a)30N

(b)24N

(c)10N

(d)4N
Solution:
The masses of two bodies are 2kg and 3kg respectively. The acceleration of A is
$$a=\frac{3-2}{3+2}g=2m/s^2$$
The net force on A will be F=ma=(2\times2)=4N.