In the motorcycle stunt called " the well of death" the track is vertical cylindrical surface of 18 m radius.take the motorcyle to be a point mass and `mu=0.8`. The minimum angular speed of the motorcycle to prevent him from sliding down should be

Options

A. 6/5 rad/s
B. 5/8 rad/s
C. 25/3 rad/s
D. none of these

Explanation

To solve the problem of determining the minimum angular speed of the motorcycle in the "well of death" stunt, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Motorcycle:**
- The motorcycle experiences gravitational force (weight) acting downwards, which is \( mg \).
- There is a normal force \( N \) acting perpendicular to the surface of the cylinder.
- There is a frictional force \( F_f \) acting upwards along the surface, which prevents the motorcycle from sliding down.

2. **Relate the Forces:**
- The frictional force can be expressed as:
\[ F_f = \mu N \]
- The gravitational force acting on the motorcycle is:
\[ F_g = mg \]

3. **Centripetal Force Requirement:**
- For the motorcycle to maintain circular motion, the net force acting towards the center of the circular path must equal the centripetal force required. The centripetal force is given by:
\[ F_c = m \omega^2 r \]
- Here, \( r \) is the radius of the cylinder and \( \omega \) is the angular speed.

4. **Set Up the Equation:**
- At the point of sliding, the frictional force must balance the gravitational force:
\[ F_f = F_g \]
- Thus, we can write:
\[ \mu N = mg \]

5. **Substituting Normal Force:**
- The normal force \( N \) can be expressed in terms of centripetal force:
\[ N = m \omega^2 r \]
- Substituting this into the equation for friction gives:
\[ \mu (m \omega^2 r) = mg \]

6. **Canceling Mass:**
- Since \( m \) appears on both sides, we can cancel it out (assuming \( m \neq 0 \)):
\[ \mu \omega^2 r = g \]

7. **Solving for Angular Speed \( \omega \):**
- Rearranging the equation gives:
\[ \omega^2 = \frac{g}{\mu r} \]
- Taking the square root:
\[ \omega = \sqrt{\frac{g}{\mu r}} \]

8. **Substituting the Values:**
- Given:
- \( g = 10 \, \text{m/s}^2 \)
- \( \mu = 0.8 \)
- \( r = 18 \, \text{m} \)
- Substitute these values into the equation:
\[ \omega = \sqrt{\frac{10}{0.8 \times 18}} \]

9. **Calculating:**
- First calculate the denominator:
\[ 0.8 \times 18 = 14.4 \]
- Now calculate:
\[ \omega = \sqrt{\frac{10}{14.4}} \approx \sqrt{0.6944} \approx 0.833 \, \text{radians/second} \]

10. **Final Answer:**
- The minimum angular speed of the motorcycle to prevent sliding down is approximately:
\[ \omega \approx 0.833 \, \text{radians/second} \]