A bob of mass $m$ is suspended by a light string of length $ L $ . It is
imparted a horizontal velocity $ v_0 $ at the lowest point A such that it completes a circle in the vertical plane.
Match Column I with Column II.
Updated On: Jul 5, 2022
$ A - p $ , $ B - q $ , $ C - s $ , $ D - r $
$ A - q $ , $ B - r $ , $ C - p $ , $ D - s $
$ A - r $ , $ B - s $ , $ C - q $ , $ D - p $
$ A - s $ , $ B - p $ , $ C - r $ , $ D - q $
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The Correct Option isC
Solution and Explanation
There are two external forces on the bob : gravity $ (mg) $ and tension $ (T) $ in the string. The tension does no work as displacement is always perpendicular to the string. Total mechanical energy $ (E) $ of the system is conserved.
If we take potential energy of the system to be zero at the lowest point $ A $ , then
At $ A $ , $ E = \frac{1}{2}mv^{2}_{0} \quad\cdots\left(i\right) $
From Newtons second law
$ T_{A} - mg = \frac{mv^{2}_{0}}{L}\quad\cdots\left(ii\right) $
where $ T_{A} $ is the tension in the string at $ A $ .
At the highest point $ C $ , to complete the circle tension in string will be minimum (zero).
At $ C $ , $ E = \frac{1}{2}mv^{2}_{C}+mg\left(2L\right)\quad\cdots\left(iii\right) $
From Newtons second law
$ mg = \frac{mv_{C}^{2}}{L}\quad\quad\left(iv\right) $
From $ \left(iv\right), v_{C} =\sqrt{gL} $ ; $ C-q $
From $ \left(iii\right) $ , $ E = \frac{1}{2}m\left(gL\right)+2mgL= \frac{5}{2}mgL \quad\left(v\right) $
Using $ \left(i\right) $ , $ \frac{1}{2}mv^{2}_{0} = \frac{5}{2}mgL $ ,
$ v_{0} = \sqrt{5\,gL} $ ;
$ \therefore A-r\quad\cdots\left(vi\right) $
At $ B $ , $ E =\frac{1}{2}mv^{2}_{B}+mg\left(L\right) $
or $ \frac{1}{2}mv^{2}_{B} = E-mg\left(L\right) $$ \frac{5}{2}mgL-mgL\quad $ (Using $ \left(v\right) $ )
$ \frac{1}{2} mv^{2}_{B} = \frac{3}{2} mgL $$ \therefore v_{B} = \sqrt{3\,gL} $ ;
$ \therefore B-s $
The ratio of kinetic energies at $ B $ and $ C $ is
$ \therefore \frac{K_{B}}{K_{C}} = \frac{\frac{1}{2}mv^{2}_{B}}{\frac{1}{2}mv^{2}_{C}} $$ = \frac{3gL}{gL} = \frac{3}{1} $ ;
$ \therefore D-p $
Work is correlated to force and the displacement over which it acts. When an object is replaced parallel to the force's line of action, it is thought to be doing work. It is a force-driven action that includes movement in the force's direction.
The work done by the force is described to be the product of the elements of the force in the direction of the displacement and the magnitude of this displacement.
Energy:
A body's energy is its potential to do tasks. Anything that has the capability to work is said to have energy. The unit of energy is the same as the unit of work, i.e., the Joule.
There are two types of mechanical energy such as; Kinetic and potential energy.
Power is the rate at which energy is transferred, conveyed, or converted or the rate of doing work. Technologically, it is the amount of work done per unit of time. The SI unit of power is Watt (W) which is joules per second (J/s). Sometimes the power of motor vehicles and other machines is demonstrated in terms of Horsepower (hp), which is roughly equal to 745.7 watts.
Power is a scalar quantity, which gives us a quantity or amount of energy consumed per unit of time but with no manifestation of direction.
