A block is mass m moves on as horizontal circle against the wall of a cylindrical room of radius R. The floor of the room on which the block moves is smooth but the friction coefficient between the wall and the block is `mu`. The block is given an initial speed `v_0`. As a function of the speed v write a. the normal force by the wall on the block. b. the frictional force by the wall and c. the tangential acceleration of the block. d. Integrate the tangential acceleration `((dv)/(dt)=v(dv)/(ds))` to obtain the speed of the block after one revoluton.

Explanation

To solve the problem step by step, we will analyze the forces acting on the block and derive the required expressions.

### Step 1: Determine the Normal Force by the Wall on the Block
The block is moving in a circular path, and the normal force (N) exerted by the wall on the block provides the necessary centripetal force for circular motion. The centripetal force required for an object of mass \( m \) moving with speed \( v \) in a circle of radius \( R \) is given by:

\[ F_c = \frac{mv^2}{R} \]

Thus, the normal force \( N \) can be expressed as:

\[ N = \frac{mv^2}{R} \]

### Step 2: Determine the Frictional Force by the Wall
The frictional force \( f \) acting on the block is given by the product of the normal force and the coefficient of friction \( \mu \):

\[ f = \mu N = \mu \left(\frac{mv^2}{R}\right) = \frac{\mu mv^2}{R} \]

### Step 3: Determine the Tangential Acceleration of the Block
The tangential acceleration \( a_t \) can be calculated using the frictional force. According to Newton's second law, the tangential acceleration is given by:

\[ a_t = \frac{f}{m} = \frac{\frac{\mu mv^2}{R}}{m} = \frac{\mu v^2}{R} \]

### Step 4: Integrate the Tangential Acceleration to Obtain Speed After One Revolution
We start with the relationship between tangential acceleration and speed:

\[ \frac{dv}{dt} = a_t = \frac{\mu v^2}{R} \]

Using the chain rule, we can write:

\[ \frac{dv}{dt} = v \frac{dv}{ds} \]

Thus, we have:

\[ v \frac{dv}{ds} = \frac{\mu v^2}{R} \]

Dividing both sides by \( v \) (assuming \( v \neq 0 \)):

\[ \frac{dv}{ds} = \frac{\mu v}{R} \]

Now, we can separate variables and integrate:

\[ \frac{dv}{v} = \frac{\mu}{R} ds \]

Integrating both sides:

\[ \int \frac{dv}{v} = \int \frac{\mu}{R} ds \]

This gives us:

\[ \ln v = \frac{\mu}{R} s + C \]

To find the constant \( C \), we use the initial condition. When \( s = 0 \), \( v = v_0 \):

\[ \ln v_0 = C \]

Substituting back into the equation:

\[ \ln v = \frac{\mu}{R} s + \ln v_0 \]

Exponentiating both sides:

\[ v = v_0 e^{\frac{\mu}{R} s} \]

### Speed After One Revolution
After one complete revolution, the displacement \( s \) is equal to the circumference of the circle, which is \( 2\pi R \):

\[ v = v_0 e^{\frac{\mu}{R} (2\pi R)} = v_0 e^{2\pi \mu} \]

### Summary of Results
a. The normal force by the wall on the block is:
\[ N = \frac{mv^2}{R} \]

b. The frictional force by the wall is:
\[ f = \frac{\mu mv^2}{R} \]

c. The tangential acceleration of the block is:
\[ a_t = \frac{\mu v^2}{R} \]

d. The speed of the block after one revolution is:
\[ v = v_0 e^{2\pi \mu} \]