Q47E - University Physics with Modern Physics

Question

Question: A small model car with mass m  travels at constant speed on the inside of a track that is a vertical circle with radius (Fig. E5.45). If the normal force exerted by the track on the car when it is at the bottom of the track (point A) is equal to 250 Mg ,how much time does it take the car to complete one revolution around the track?

 

Figure E5.45

Step-by-Step Solution

Answer

Answer:

 

The time taken by car to complete one revolution is 3.66s.

 

Step 1: Identification of given data

The given data can be listed below as,

• The mass of the car is,m .
• The radius of the track is,r = 5.00m .
• The normal force at the bottom is,N = 250 mg .

 

Step 2: Significance of normal force

The normal force is the force of contact applied by a surface on an object. The normal force only occurs when the two surfaces are in contact.

Step 3: Determination of time taken by car to complete one revolution.

The expression for the normal force using newton’s second law in a circular motion can be expressed as,

 $N-m g=\frac{m v^2}{r}$

Here N is the normal force, m  is the mass of the car, g is the acceleration due to gravity, v  is the velocity, and  r is the radius of the track.

 

Substitute 250 mg forN ,  9.8 m/s ²for g, 500 m for r in the above equation.  

 $$\begin{gathered} 2.50 mg-m g=\frac{m v^2}{5.00 \text{m}} \\ m g (2.50-1)=\frac{m v^2}{5.00 \text{m}} \\ 1.5 g\times 5.00 \text{m}=v^2 \\ v=\sqrt{1.5\times 9.8 {\text{m/s}}^{\text{2}}\times 5.00 \text{m}} \\ =8.57 \text{m/s} \end{gathered}$$

 

Hence, the velocity is,8.57 m/s .

 

The expression for the time period can be expressed as,

 $T=\frac{2 \pi r}{v}$

 

Substitute 500 m for r ,8.57  for v , 3.14 for in the above equation.

 $$\begin{gathered} T=\frac{2\times 3.14\times 5.00 \text{m}}{8.57 \text{m/s}} \\ =3.66 sec \end{gathered}$$

 

Hence, the required time is, 3.66 sec.