Q20P - Fundamentals Of Physics

Question

A pendulum consists of a$\text{2.0kg}$stone swinging on a string of $\text{4.0 m}$negligible mass. The stone has a speed of $\text{8.0m/s}$when it passes its lowest point.

1. What is the speed when the string is at${60}^0$to the vertical? 
2. What is the greatest angle with the vertical that the string will reach during the stone’s motion? 
3. If the potential energy of the pendulum–Earth system is taken to be zero at the stone’s lowest point, what is the total mechanical energy of the system?

Step-by-Step Solution

Answer

1. The speed of the pendulum when the string is at ${60}^o$its vertical is $\text{5m/s}$
2. The greatest angle with the vertical that the string will reach during the stone’s motion is ${\text{794}}^o$
3. The total mechanical energy of the system is $\text{64J}$

Step 1: Given

1. A mass of pendulums, $\text{M=2.0kg}$
2. Length of string, $L=40m$
3. Velocity at its lowest point is,  $\text{V=80 m/s}$

Step 2: Determining the concept

Using the energy conservation law, find the velocity at the lowest point. So, according to the conditions in the problem, find the velocity at the lowest point and the greatest angle of string also. According to the law of energy conservation, energy can neither be created nor destroyed.

 Formula:

$$\begin{gathered} Energy=PE+KE \\ =\frac{1}{2}m{v}^2+mgh \end{gathered}$$

where, KE is kinetic energy, PE is potential energy, m is mass, v is velocity, g is an acceleration due to gravity and h is height.

Step 3: (a) Determining the speed of the pendulum when the string is at to the vertical

From the figure, 

$\begin{array}{l}h=L-L\left(\cos \theta \right)\\ h=L\left(1-\left(\cos \theta \right)\right)\end{array}$

The total energy of the system is,

 $$\begin{gathered} E_{total}=K{E}_{lowest} \\ =P{E}_{top} \\ ={\left(KE+PE\right)}_{Medieval} \end{gathered}$$

 At the top, the object has only potential energy, and at the lowest point, it has only kinetic energy. Therefore,

$$\begin{gathered} E=\frac{1}{2}m{v}^2 \\ E=\frac{1}{2}\left(2\right){\left(8\right)}^2 \\ E=64\text{J} \end{gathered}$$

From the law of conservation of energy, it is clear that,

 $$\begin{gathered} E=KE+PE \\ 64\text{J}=\frac{1}{2}m{v}_0^2+mgh \\ 64\text{J}=\frac{1}{2}m{v}_0^2+mg\left(L\left(1-\left(\cos \theta \right)\right)\right) \\ 64\text{J}=\frac{1}{2}m{v}_0^2+mgL-mgL\cos \theta \\ 64\text{J}=\frac{1}{2}\left(2.0\right)v_0^2+\left(2.0\right)\left(9.8\right)\left(4.0\right)-\left(2.0\right)\left(9.8\right)\left(4.0\right)\cos 60 \\ 64\text{J}=v_0^2+78.4-39.2 \\ {v}_0^2=24.8 \\ {v}_0=4.97 \\ =5\text{m/s} \end{gathered}$$

Hence, the speed of pendulum when the string is at  to the vertical is $\text{5m/s.}$

Step 4: (b) Determining the greatest angle with the vertical that the string will reach during the stone’s motion.

At the greatest angle, the velocity of stone will be zero. So, it has only potential energy.

So, 

$$\begin{gathered} E=\frac{1}{2}m{v}^2+mgh \\ E=0+mgh \\ E=mgL\left(1-\left(\cos \theta \right)\right) \\ 64=\left(2.0\right)\left(9.8\right)\left(4.0\right)\left(1-\cos \theta \right) \\ \frac{64}{78.4}=\left(1-\cos \theta \right) \\ \cos \theta =1-0.8163 \\ \theta ={\cos }^{-1}\left(0.1836\right) \\ \theta =79.4° \end{gathered}$$

Hence, the greatest angle with the vertical that the string will reach during stone’s motion is $79.4°.$

Step 5: (c) Determining the total mechanical energy of the system

As, it is already calculated, the energy of the system is $\text{64 J}$ , and the law of conservation of energy says that the energy of the system will remain constant. So, the total mechanical energy is $\text{64 J}$.

 

Hence, total mechanical energy of system is $\text{64 J}$