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.
Diagram of the pulley
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.
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:
Diagram of the system
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$$2Tcos\theta=Mg$$$$2(M+m)g\times\frac{10}{\sqrt{1300}}=Mg$$$$2(M+m)\times\frac{1}{\sqrt{13}}=M$$$$ 2(M+m)=\sqrt{13}M$$$$2m=M(\sqrt{13}-2)$$$$m=\frac{M\times(\sqrt{13}-2)}{2}=0.9M$$
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
$$ma=2mg$$
$$\Rightarrow a=g/2$$
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.
