Question

11. If f'(x) and g'(x) exist for all x E[a, b], and if g'(x) does not vanish anywhere on ]a, b[. Then prove that for
some c between a and b.
$\frac{f(c) - f(a)}{g(b) - g(c)} = \frac{ {f}^{1}(c) }{ {g}^{1}(c) } \\ \\ (here \: 1 \: means \: dash)$

​

Answer 1

Answer:

1. Weight of the person (mg) acts in a downward direction.

2. A normal force acts perpendicular to the surface (towards the center).

3. A centripetal force acts (due to circular motion) towards the centre with magnitude mω2r where m is the mass, ω is angular velocity and r is the distance from the axis.

4. A frictional force acts opposite in direction to the downward force (mg).

Now, we can conclude that the normal force is equal to the centripetal force. Therefore,

N=mω2r.

We know that a frictional force is given the expression:

f=μN

Keeping the value of N, we get

f=μmω2r.

This frictional force is balancing the download force i.e., the weight of the body. Upon equating the two forces, we get:

mg=μmω2r.

Cancelling the m on both sides:

g=μω2r.

We are asked to find the minimum angular velocity ω,

So,

ω=gμr−−−√

So, the correct answer is “Option C”.

Note:

Note:The condition for minimum velocity is when the friction completely balances mg; beyond this minimum angular velocity, the person certainly continues the circular motion and has no chance of falling down. If the angular velocity is lower than the minimum value, gravity dominates and the person falls.