Askiitians Tutor Team
Last Activity: 5 Months ago
To determine the minimum coefficient of friction \( u \) required for a block of mass \( m \) to maintain complete vertical circular motion inside a rotating cylinder, we need to analyze the forces acting on the block as it moves along the inner surface of the cylinder. The key here is to ensure that the block does not slip and can complete the circular path.
Understanding the Forces at Play
When the block is at the top of the circular path, the forces acting on it include:
- The gravitational force acting downward, \( F_g = mg \).
- The normal force exerted by the cylinder's surface, \( N \), acting perpendicular to the surface.
At the top of the circular motion, the centripetal force required to keep the block moving in a circle is provided by the combination of the gravitational force and the normal force. The equation for centripetal force \( F_c \) can be expressed as:
1. Centripetal Force Equation:
At the top of the circle, the centripetal force is given by:
\[ F_c = \frac{mv^2}{R} \]
Where \( v \) is the tangential velocity of the block. At the top, we have:
\[ N + mg = \frac{mv^2}{R} \]
Velocity of the Block
Since the cylinder is rotating with an angular velocity \( W \), the tangential velocity \( v \) of the block can be expressed as:
\[ v = WR \]
Substituting \( W = \sqrt{\frac{5g}{R}} \) into the equation gives:
\[ v = R \sqrt{\frac{5g}{R}} = \sqrt{5gR}
\]
Substituting Back into the Centripetal Force Equation
Now substituting \( v \) back into the centripetal force equation:
\[ N + mg = \frac{m(\sqrt{5gR})^2}{R} \]
Which simplifies to:
\[ N + mg = \frac{m(5gR)}{R} = 5mg
\]
From this, we can isolate the normal force \( N \):
\[ N = 5mg - mg = 4mg
\]
Frictional Force Requirement
For the block to maintain its circular motion without slipping, the frictional force must be sufficient to provide the necessary centripetal force when the block is at the bottom of the circular path. The frictional force \( F_f \) can be expressed as:
\[ F_f = uN
\]
At the bottom of the circular path, the equation for centripetal force becomes:
\[ N - mg = \frac{mv^2}{R}
\]
Substituting \( N \) from earlier:
\[ 4mg - mg = \frac{m(5gR)}{R}
\]
Which simplifies to:
\[ 3mg = 5mg
\]
Calculating the Minimum Coefficient of Friction
Setting the frictional force equal to the required centripetal force gives:
\[ u(4mg) = 3mg
\]
Solving for \( u \) yields:
\[ u = \frac{3mg}{4mg} = \frac{3}{4}
\]
Final Result
Thus, the minimum coefficient of friction \( u \) required for the block to complete its vertical circular motion without slipping is:
\[ u \geq \frac{3}{4}
\]
This means that the friction between the block and the cylinder must be at least 0.75 for the block to successfully navigate the circular path without losing contact. Understanding these dynamics helps in analyzing similar problems in rotational motion and friction.
