Question

A small block is placed on the top of a frictionless hemisphere of radius r. The hemisphere is placed on a horizontal floor as shown in the figure. The hemisphere is given a constant acceleration $a_0$ towards the left.

At what angular displacement relative to the center of the hemisphere will the block lose contact with the hemisphere.

Answer 1

θ = arccos[(6g^2 + a_0 sqrt(5g^2 + a_0^2)) / (9g^2 + a_0^2)]

Answer 2

To determine the angular displacement at which the block loses contact with the hemisphere, we will analyze the forces acting on the block in the non-inertial frame of reference of the hemisphere.

Let:

• m be the mass of the block.
• r be the radius of the hemisphere.
• g be the acceleration due to gravity.
• a_0 be the constant acceleration of the hemisphere towards the left.
• θ be the angular displacement of the block from the vertical (measured from the top of the hemisphere).

1. Free Body Diagram and Equations of Motion in the Non-Inertial Frame:

In the frame of reference of the accelerating hemisphere, a pseudo force F_pseudo = ma_0 acts on the block in the direction opposite to the hemisphere's acceleration, i.e., towards the right.

Consider the block at an angular position θ from the vertical. We resolve forces along the radial and tangential directions.

• Radial Direction (towards the center of the hemisphere):

  The forces acting radially are the component of gravity mg cosθ, the component of the pseudo force ma_0 sinθ, and the normal force N (acting outwards). The net radial force provides the centripetal acceleration mv^2/r. Taking the inward radial direction as positive: mg cosθ + ma_0 sinθ - N = mv^2/r When the block loses contact, the normal force N = 0. So, the condition for losing contact is: mg cosθ + ma_0 sinθ = mv^2/r (Equation 1)

• Tangential Direction (perpendicular to the radius, along the direction of motion):

  The forces acting tangentially are the component of gravity mg sinθ and the component of the pseudo force ma_0 cosθ. These forces cause the block to accelerate tangentially. Taking the tangential direction downwards as positive: mg sinθ + ma_0 cosθ = m (dv/dt)

2. Work-Energy Theorem to find v^2:

We can use the work-energy theorem in the non-inertial frame. The work done by all forces (including pseudo forces) equals the change in kinetic energy. Initial state: Block at rest on top, v_initial = 0. Final state: Block at angle θ with velocity v.

• Work done by gravity (W_g):

  The vertical distance fallen by the block is h = r - r cosθ = r(1 - cosθ). W_g = mg * h = mgr(1 - cosθ)

• Work done by pseudo force (W_pseudo):

  The pseudo force ma_0 acts horizontally to the right. The horizontal displacement of the block is x = r sinθ (to the right). W_pseudo = F_pseudo * x = ma_0 (r sinθ)

• Work done by normal force: W_N = 0 (as N is perpendicular to displacement).

According to the work-energy theorem: W_g + W_pseudo = ΔK mgr(1 - cosθ) + ma_0 r sinθ = (1/2)mv^2 - 0 Divide by m: gr(1 - cosθ) + a_0 r sinθ = (1/2)v^2 v^2 = 2gr(1 - cosθ) + 2a_0 r sinθ (Equation 2)

3. Substitute v^2 into the Contact Condition:

Substitute Equation 2 into Equation 1: mg cosθ + ma_0 sinθ = (m/r) * [2gr(1 - cosθ) + 2a_0 r sinθ] Divide by m: g cosθ + a_0 sinθ = (1/r) * [2gr(1 - cosθ) + 2a_0 r sinθ] g cosθ + a_0 sinθ = 2g(1 - cosθ) + 2a_0 sinθ

Rearrange the terms to solve for cosθ and sinθ: g cosθ + 2g cosθ = 2g + 2a_0 sinθ - a_0 sinθ 3g cosθ = 2g + a_0 sinθ 3g cosθ - a_0 sinθ = 2g (Equation 3)

4. Solve for θ:

To solve this equation, we can express sinθ in terms of cosθ using sinθ = sqrt(1 - cos^2θ) (assuming θ is in the first quadrant, which is physically relevant for sliding down). 3g cosθ - 2g = a_0 sqrt(1 - cos^2θ) Square both sides: (3g cosθ - 2g)^2 = a_0^2 (1 - cos^2θ) g^2 (3 cosθ - 2)^2 = a_0^2 (1 - cos^2θ) g^2 (9 cos^2θ - 12 cosθ + 4) = a_0^2 - a_0^2 cos^2θ 9g^2 cos^2θ - 12g^2 cosθ + 4g^2 = a_0^2 - a_0^2 cos^2θ Rearrange into a quadratic equation in cosθ: (9g^2 + a_0^2) cos^2θ - 12g^2 cosθ + (4g^2 - a_0^2) = 0

Let x = cosθ. Using the quadratic formula x = [-B ± sqrt(B^2 - 4AC)] / 2A: cosθ = [12g^2 ± sqrt((-12g^2)^2 - 4(9g^2 + a_0^2)(4g^2 - a_0^2))] / [2(9g^2 + a_0^2)] cosθ = [12g^2 ± sqrt(144g^4 - 4(36g^4 - 9g^2a_0^2 + 4g^2a_0^2 - a_0^4))] / [2(9g^2 + a_0^2)] cosθ = [12g^2 ± sqrt(144g^4 - 4(36g^4 - 5g^2a_0^2 - a_0^4))] / [2(9g^2 + a_0^2)] cosθ = [12g^2 ± sqrt(144g^4 - 144g^4 + 20g^2a_0^2 + 4a_0^4)] / [2(9g^2 + a_0^2)] cosθ = [12g^2 ± sqrt(4a_0^2(5g^2 + a_0^2))] / [2(9g^2 + a_0^2)] cosθ = [12g^2 ± 2a_0 sqrt(5g^2 + a_0^2)] / [2(9g^2 + a_0^2)] cosθ = [6g^2 ± a_0 sqrt(5g^2 + a_0^2)] / [9g^2 + a_0^2]

We must choose the sign that corresponds to the physical scenario. From Equation 3, 3g cosθ = 2g + a_0 sinθ. Since a_0 > 0 and sinθ > 0 (for 0 < θ < 90°), it implies 3g cosθ > 2g, or cosθ > 2/3. Let's check the two solutions for cosθ. The positive sign + will result in a larger value of cosθ (hence smaller θ). The negative sign - will result in a smaller value of cosθ (hence larger θ). For the block to slide down and then lose contact, θ must be positive. If a_0 = 0, cosθ = 6g^2 / 9g^2 = 2/3. This is the standard result. If a_0 > 0, the pseudo force ma_0 assists the block in sliding down. This means it will lose contact at an earlier angle (smaller θ, larger cosθ) compared to a_0 = 0. Therefore, we choose the + sign to get cosθ > 2/3.

Thus, the angular displacement θ is given by: cosθ = (6g^2 + a_0 sqrt(5g^2 + a_0^2)) / (9g^2 + a_0^2) θ = arccos[(6g^2 + a_0 sqrt(5g^2 + a_0^2)) / (9g^2 + a_0^2)]