13.13 Conservation of mechanical energy  (Page 2/2)

Here, we shall work an example that we had previously analyzed with the help of "work-kinetic energy" theorem. This will help us to understand (i) what are the elements of ideal mechanical process (ii) how conservation principle works and (iii) how does it help simplify analysis.

Example

Problem : A small block of 0.1 kg is released from a height 5 m as shown in the figure. The block following a curved path transitions to a linear horizontal path and hits the spring fixed to a wedge. If no friction is involved and spring constant is 1000 N/m, find the maximum compression of the spring.

Motion of a block

Solution : Here block, curved incline and spring form isolated system. Earth is implicit element of the system. We may argue why incline and why not horizontal surface? Curved incline (its shape) determines the component of gravity driving block downward. Hence, it is included in the system. Horizontal surface also applies force on block, but in normal direction to the motion. Now, as friction is not involved, the horizontal surface is not involved in the exchange of energy.

Two of the elements i.e. block and spring are subjected to motion. Application of conservation principle, thus, involves more than one object unlike analysis based on laws of motion. As friction is not involved, we can apply the conservation of mechanical energy. Now, initial energy of the system is :

$$\begin{array}{l}\Rightarrow E_i=K+U=0+mgh=mgh=0.1x10x5=5J\end{array}$$

When the compression in the spring is maximum, kinetic energies of both block and spring are zero. Let "x" be the maximum compression,

$$\begin{array}{l}\Rightarrow E_f=K+U=0+\frac{1}{2}k{x}^2=\frac{1}{2}k{x}^2\\ \Rightarrow E_f==0.5x1000x^2=500x^2\end{array}$$

According to conservation of mechanical energy :

$$\begin{array}{l}\Rightarrow 500x^2=5\\ \Rightarrow x=0.1m=10\mathrm{cm}\end{array}$$

Vertical circular motion

Study of this vertical motion was incomplete earlier as we were limited by constant force system and linear motion in most of the cases. On the other hand, vertical circular motion involved variable force and non-linear path. The analysis of the situation is suited to the energy analysis. Let us consider that a small object of mass "m" is attached to a string and is whirled in a vertical plane.

Force analysis at the highest point gives the minimum speed required so that string does not slack. It is important to understand that this is the most critical point, where string might slack. The forces in y-direction :

Vertical circular motion

$$\begin{array}{l}\sum F_y=mg+T=\frac{m{v}^2}{r}\end{array}$$

For the limiting case, when T = 0,

$$\begin{array}{l}\Rightarrow mg=\frac{m{v}^2}{r}\\ \Rightarrow v=\surd \left(gr\right)\end{array}$$

This expression gives the limiting value of the speed of particle in vertical circular motion. If the speed of the particle is less than this limiting value, then the particle will fall. Once this value is known, we can find speed of the particle at any other point along the circular path by applying conservation of mechanical energy. Here, we shall find speed at two other positions - one at the bottom (C) and other at the middle point (B or D).

Vertical circular motion

In order to analyze mechanical energy, we need to have a reference zero gravitational potential energy. For this instant case, we consider the horizontal level containing point "C" as the zero reference for gravitational potential energy. Now, mechacnial energy at points "A", "C" and "D" are :

$$\begin{array}{l}\Rightarrow E_A=K+U=\frac{1}{2}m{v_A}^2+mgx2r=\frac{1}{2}mgr+2mgr=\frac{5}{2}mgr\\ \Rightarrow E_C=K+U=\frac{1}{2}m{v_C}^2+0=\frac{1}{2}m{v_C}^2\\ \Rightarrow E_D=K+U=\frac{1}{2}m{v_D}^2+mgr\end{array}$$

From first two equations, we have :

$$\begin{array}{l}\Rightarrow v_C=\surd \left(5gr\right)\end{array}$$

From last two equations, we have :

$$\begin{array}{l}\Rightarrow v_D=\surd \left(3gr\right)\end{array}$$

This shows how energy analysis can be helpful in situations where force on the object is not constant and path of motion is not straight. However, this example also points out one weak point about energy analysis. For example, the initial condition for minimum speed at the highest can not be obtained using energy analysis and we have to resort to force analysis. What it means that if details of the motion are to be ascertained, then we have to take recourse to force analysis.

Example

Problem 2: The ball is released from a height “h” along a smooth path shown in the figure. What should be the height “h” so that ball goes around the loop without falling of the track ?

Vertical circular motion

Solution : The path here is smooth. Hence, friction is absent. No external force is working on the “Earth – ball” system. The only internal force is gravity, which is a conservative force. Hence, we can apply conservation of mechanical energy.

The most critical point where ball can fall off the track is “B”. We have seen that the ball need to have a minimum speed at "B" as given by :

$$v_B=\sqrt{gr}$$

Corresponding to this requirement at “B”, the speed of the ball can be obtained at bottom point “C” by applying conservation of mechanical energy. The speed of the ball as worked out earlier is given as :

$$v_C=\sqrt{5gr}$$

In order to find the height required to impart this speed to the ball at the bottom, we apply law of conservation of energy between “A” and “C”.

$$K_i+U_i=K_f+U_f$$

Considering ground as zero reference gravitational potential, we have :

$$\Rightarrow 0+mgh=\frac{1}{2}m{v}_C^2+0$$

$$\Rightarrow v_C^2=\frac{2mgh}{m}=2gh$$

$$\Rightarrow v_C=\sqrt{2gh}$$

Equating two expressions of the speed at point “C”, we have :

$$\Rightarrow 5gr=2gh$$

$$\Rightarrow h=\frac{5r}{2}$$