Question

Two bodies of mass m and 4m are attached with a string as
shown in the figure. The body of mass m hanging from a string of
length L is executing oscillations of angular amplitude thetanot while the
other body is at rest. Find out the minimum coefficient of friction
for the body of mass 4m to be at rest.

Plss answer fast ​

Answer 1

Answer:

The answer will be (3 -costheta)/4

Explanation:

According to the figure we can see that the length of the string is L and mass of both the bodes are m and 4 m respectively.

Now the mass m is making angle of thetanot .

Therefore, costheta = h /L

              => h = Lcostheta

now for h'

h' = L-h = L- Lcostheta

Now for total energy,

for point 1

mg(h') = 1/2 mv^2

=> v = √2gL(1-costheta)

Now for the tension is the wire,

Tmax - mg = mv^/L

=>Tmax = mg (3 -costheta)

Now , μ4mg >= T max

           => μ4mg = mg (3 -costheta)

            => μ = (3 -costheta)/4