Interesting physics problem about a rope attached to an inclined plane

Question

After many attempts, I don't really understand how to solve this physics problem:

A rope of lenght l=20 cm is attached to an inclined plane (plane's angle of displacement is α = 30°). The other end of the rope is attached to a small object that is capable of circular motion on the inclined plane. The object starts moving from the lowest position such that the velocity on the highest position of its trajectory is v=3 m/s. Find:

a) the initial velocity of the object if on the highest position the tension of the rope is 2 times smaller than on the initial position.

b) the coefficient of friction μ between the object and the inclined plane

c) the full distance the object moved before stopping, if after the object rotates 5/4 times, the rope breaks and the object stays on the inclined plane before it stops.

Answer 1

Please draw a picture of the inclined plane from the side, making it appear as a triangle.

You cannot draw the circular trajectory, but you can identify the highest and lowest point.

Let

l = 0.2 m be the length of the rope,

α = 30° be the angle of the plane from the horizontal,

m (unknown) be the mass of the object,

v = 3 m/s be its velocity at the highest point,

v₀ (to be calculated) be its velocity at the initial lowest point,

T (unknown) be the tension of the rope at the highest point,

T₀ (unknown) be the tension of the rope at the initial lowest point,

g = 9.81 m/s² be gravitational acceleration.

The object is subject to three forces:

- gravitational force with magnitude mg directed downward,

- tension directed from the object towards the center of rotation,

- force of friction tangential to the trajectory.

(You cannot draw the last one in your picture, but we will not need that).

We are given

T₀ = 2T (1)

a)

The object's acceleration can be decomposed into two orthogonal components:

- centripetal acceleration directed towards the center, and

- tangential acceleration.

(We will need to talk only about the centripetal acceleration.)

The centripetal acceleration is due to centripetal force, which comes from the projection

of the three forces on the line of the string.

The force of friction, being perpendicular to the string, contributes nothing to the centripetal force.

The force of gravity contributes mgsin(α).

At the highest point that contribution is in the same direction as tension, while

at the lowest point that contribution is opposite to tension.

Newton's Second Law applied to the highest point is

mv²/l = T + mgsin(α) (2)

Applied to the lowest point

mv₀²/l = T₀ - mgsin(α) (3)

Use (1) and (2) to eliminate T₀ from (3)

mv₀²/l = 2(mv²/l - mgsin(α)) - mgsin(α)

From that

v₀ = √(2v² - 3glsin(α)) (4)

Substituting actual numbers into (4)

v₀ = √(2×3² - 3×0.2×sin(30°)) = 4.2 m/s

b)

Use conservation of energy.

Let the bottom point be the reference point for potential energy.

Then at the bottom, the object has potential energy 0, and kinetic energy mv₀²/2.

At the top, the object has potential energy 2mglsin(α), and kinetic energy mv²/2.

The energy at the bottom gets converted to the energy at the top plus the work of friction, W:

W = Fs,

where F is the force of friction and s is the length traversed from the lowest point to the highest:

s = πl

Then

F = Nμ, where N is the normal force, which is due to gravity alone.

N = mgcos(α)

Thus

W = mgcos(α)μπl

Conservation of energy is then expressed by

mv₀²/2 = 2mglsin(α) + mv²/2 + mgcos(α)μπl

From that

μ = (v₀² - 4glsin(α) - v²)/2πglcos(α)

From (4)

μ = (2v² - 3glsin(α) - 4glsin(α) - v²)/2πglcos(α)

μ = (v² - 7glsin(α))/2πglcos(α) (5)

Substituting actual numbers

μ = (3² - 7×9.81×0.2×sin(30°))/(2×π×9.81×0.2×cos(30°)) = 0.2

c)

The rope breaks when it is horizontal, and the object continues upward until it stops after some distance h.

The full distance traversed is (2πl)×(5/4) + h = 5πl/2 + h

To calculate h we again use conservation of energy.

At the stopping point the object has kinetic energy 0 and potential energy mg(l+h)sin(α).

We already calculated the energy at the highest point as 2mglsin(α) + mv²/2.

That energy gets converted to the energy at the stopping point plus work of friction over the distance

3πl/2 + h.

The equation of conservation of energy is

2mglsin(α) + mv²/2 = mg(l+h)sin(α) + mgcos(α)μ(3πl/2 + h)

Simplify into

4glsin(α) + v² = 2g(l+h)sin(α) + gcos(α)μ(3πl + 2h)

From (5) we can replace

gcos(α)μ = (v² - 7glsin(α))/2πl

to get

4glsin(α) + v² = 2g(l+h)sin(α) + (v² - 7glsin(α))(3πl + 2h)/2πl

Express h

h = (25glsin(α) - v²)πl/(glsin(α)(4π - 14) + 2v²)

Substituting actual numbers

h = (25×9.81×0.2×sin(30°) - 3²)×π×0.2/(9.81×0.2×sin(30°)(4×π - 14) + 2×3²) = 0.59

The full distance is

(2×π×0.2)×(5/4) + 0.59 = 2.16 m