7. A smooth ring A of mass m can slide on a fixed horizontal
rod. A string tied to the ring passes over a fixed pulley
B and carries a block C of mass M (=2 m) as shown in
figure (5-W7). At an instant the string between the ring
and the pulley makes an angle 0 with the rod.
(a) Show
that, if the ring slides with a speed u, the block descends
with speed v cos 0.
(b) With what acceleration will the
ring start moving if the system is released from rest with
0 = 30° ?
rod. A string tied to the ring passes over a fixed pulley
B and carries a block C of mass M (=2 m) as shown in
figure (5-W7). At an instant the string between the ring
and the pulley makes an angle 0 with the rod.
(a) Show
that, if the ring slides with a speed u, the block descends
with speed v cos 0.
(b) With what acceleration will the
ring start moving if the system is released from rest with
0 = 30° ?
Answers & Comments
Answer:
(a) To show that the block C descends with speed v cos θ, we can consider the forces acting on the block C and the ring A.
1. Forces on Block C:
- The weight of block C: M * g (where g is the acceleration due to gravity).
- Tension in the string: T (same tension throughout the string).
- Tension component along the vertical direction: T * cos θ (since the angle between the string and the vertical is θ).
- Net force on block C along the vertical direction: M * g - T * cos θ.
Using Newton's second law, we have:
M * a = M * g - T * cos θ ...........(1)
where 'a' is the acceleration of the block C along the vertical direction.
2. Forces on Ring A:
- Tension in the string: T (same tension throughout the string).
- Friction force on the ring A: f (opposing its motion).
- Net force on the ring A along the horizontal direction: f.
Using Newton's second law for the ring A, we have:
m * u = f ...........(2)
where 'u' is the speed of the ring A.
Since the ring A is sliding smoothly, the friction force f is given by:
f = μ * m * g ...........(3)
where 'μ' is the coefficient of friction between the ring and the rod.
Now, the length of the string between the ring and the pulley remains constant, so the displacement of the ring A is equal to the displacement of the block C.
Distance traveled by block C = Distance traveled by ring A
(M * v * cos θ) = (u - v) ...........(4)
where 'v' is the speed of the block C.
Now, we can relate 'u' and 'v' using equations (1), (2), (3), and (4).
From equation (1), we have:
a = g - (T * cos θ) / M ...........(5)
From equation (2), we have:
u = μ * g ...........(6)
From equation (3), we have:
f = μ * m * g ...........(7)
From equation (4), we have:
v = (u * cos θ) / (1 + cos θ) ...........(8)
Substitute equation (6) and equation (7) into equation (2):
m * u = μ * m * g
Solve for 'u':
u = μ * g ...........(9)
Substitute equation (9) into equation (8):
v = (μ * g * cos θ) / (1 + cos θ) ...........(10)
Now, we can rewrite equation (5) as:
a = g - (μ * m * g * cos θ) / (1 + cos θ) ...........(11)
So, we have shown that the block C descends with speed v cos θ and the acceleration of the ring A is given by g - (μ * m * g * cos θ) / (1 + cos θ).
(b) When the system is released from rest with θ = 30°, the acceleration of the ring A is:
a = g - (μ * m * g * cos 30°) / (1 + cos 30°)
Substitute the values:
a = g - (μ * m * g * √3) / (1 + √3)
Now, we can calculate the coefficient of friction (μ) using the fact that the ring starts moving when the friction force equals the limiting friction force:
μ = tan(θ)
μ = tan(30°)
μ = √3
Now, substitute the value of μ into the equation for 'a':
a = g - (√3 * m * g * √3) / (1 + √3)
a = g - (3 * m * g) / (1 + √3)
So, the acceleration of the ring A when the system is released from rest with θ = 30° is:
a = g - (3 * m * g) / (1 + √3)
Explanation: