Without a starting velocity, a point-pike mass m is put at the edge of a semispherical shell with radius R. The deviation's initial angle is 90 degrees.

Question

Without a starting velocity, a point-pike mass m is put at the edge of a semispherical shell with radius R. The deviation's initial angle is 90°. The mass oscillates, and after one period, we learn that the maximum deviation angle is 90°–α, where the angle α << 1 (rad). The acceleration of gravity is equal to g.

a) Using the system parameters m, R, g, calculate the average viscous friction force Ffr acting on the mass during one period of oscillation.

b) Assume that there is no friction. Using the system parameters of mass m, gravity acceleration g, and radius R, find the formula that establishes the relationship between the kinetic energy K of the mass and θ, which is the angle of departure from the vertical direction. Using your derived formula for half period, plot the kinetic K(θ) and potential U(θ) energy vs. θ. Analytically find the areas below the graph.

Answer 1

I am going to measure all angles in radians.

That will make the arc with angle θ to have the length Rθ, for any angle θ.

Therefore, instead of 90°– α I will write π/2 - α.

I am assuming that "point-pike" is to be "point-like", which means that the mass has

moment of inertia 0.

I am assuming that the assumption α << 1 allows us to approximate sin(θ) ≈ θ for any θ ≤ α.

For simplicity I am going to write F instead of Ffr.

a)

After the first half period, the mass reaches the other side of the shell at a maximum

deviation π/2-β, for some β < α.

We have two unknowns -- F and β, and we are going to write two equations to solve.

The two equations are going to describe the loss of energy due to friction

during the first and second half-period.

During the first half period the mass travels over an arc of angle π-β;

therefore the force of friction will perform work equal

R(π-β)F

This work equals the loss of potential energy at the end of the first half period, which is

mgRsin(β) ≈ mgRβ.

So the work-energy equivalence is expressed by

R(π-β)F = mgRβ

Simplify into

(π-β)F = mgβ (1)

So

F = mgβ/(π-β) (2)

During the second half period the mass travels over an arc of angle π-β-α;

therefore the force of friction will perform work equal

R(π-β-α)F

This work equals the loss of potential energy at the end of the first half period,

which is

mgRsin(α) - mgRsin(β) ≈ mgR(α-β).

So the work-energy equivalence is expressed by

R(π-β-α)F = mgR(α-β)

Rewrite it as

(π-β)F - αF = mgα - mgβ

Using (1)

mgβ - αF = mgα - mgβ

So

F = mg(2β - α)/α (3)

Substituting from (2)

mgβ/(π-β) = mg(2β - α)/α

Simplify into

β/(π-β) = (2β - α)/α

αβ = 2πβ - 2β² - πα + αβ

2β² - 2πβ + πα = 0

β = (2π ± √(4π² - 8πα))/4 = (π ± √(π² - 2πα))/2

Only the solution where β < π/2 makes physical sense, so

β = (π - √(π² - 2πα))/2 (4)

Plugging this into (2) and simplifying

F = mg(π - √(π² - 2πα))/(π + √(π² - 2πα)) (5)

Equation (5) is the formula for the force of friction derived under the approximation

sin(α) ≈ α and sin(β) ≈ β.

Let's see if we can further simplify (5) taking advantage of α << 1.

I will show that as α->0, β->α/2.

Use equation (4) and consider the ratio

β/α = ((π/α) - √((π/α)² - 2(π/α)))/2

= ((π/α) - √((π/α)² - 2(π/α) + 1 - 1))/2

= ((π/α) - √((π/α) - 1)² - 1))/2

As α->0,

the quotient π/α->∞,

so ((π/α) - 1)²->∞,

(π/α) - 1)² - 1 becomes increasingly equal (π/α) - 1)²

Therefore

lim(β/α) = lim(((π/α) - √((π/α) - 1)² - 1))/2)

= lim(((π/α) - √((π/α) - 1)²)/2)

= lim(1/2) = 1/2

Therefore when α << 1 we can approximate β ≈ α/2.

The interesting thing is that if we substitute β=α/2 into equation (3) we would get F = 0.

That means that the approximation β=α/2 is too inaccurate for calculating F.

Nevertheless, we can get a simplification of (5) if we use β=α/2 in (4):

α/2 = (π - √(π² - 2πα))/2

From that

√(π² - 2πα) = π - α

Then we can simplify (5) into

F = mgα/(2π-α)

b)

Consider the mass at an angle θ with respect to the vertical.

The range of θ is -π/2 ≤ θ ≤ π/2

At that angle θ calculate the potential energy with respect to the bottom of the shell.

The mass is at height R - Rcos(θ).

So the potential energy is

U(θ) = mgR(1 - cos(θ))

Now derive the formula for the kinetic energy K(θ).

At the top, where θ = π/2

U(π/2) = mgR

and we are given that

K(π/2) = 0

So by conservation of energy

K(θ) = mgR - mgR(1 - cos(θ)) = mgRcos(θ)

The area below the graph of K(θ) is

∫K(θ)dθ = ∫mgRcos(θ)dθ = mgRsin(θ) + K(0) = mgRsin(θ) + MgR = mgR(sin(θ) + 1)