Question

A small block is placed on the top of a frictionless hemisphere of radius $r$. The hemisphere is fixed on a horizontal floor as shown in the figure. The block is given a vanishingly small horizontal velocity. As a result the block begins to slide down the hemisphere.

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

(b) What is horizontal displacement of the block when it hits the floor.

Answer 1

(a) $\cos^{-1}(2/3)$, (b) $\frac{r}{27}(9\sqrt{5} + 10\sqrt{15})$

Answer 2

The problem involves a small block sliding down a frictionless hemisphere. We need to determine two quantities:

(a) The angular displacement at which the block loses contact with the hemisphere.

(b) The horizontal displacement of the block when it hits the floor after losing contact.

Let m be the mass of the block and r be the radius of the hemisphere. The block starts from the top of the hemisphere with a vanishingly small horizontal velocity, meaning its initial velocity is approximately zero.

Part (a): Angular displacement at which the block loses contact.

Let θ be the angular displacement of the block from the vertical (the top of the hemisphere) when it loses contact.

1. Conservation of Mechanical Energy: The block starts at a height r from the center of the hemisphere. Let's set the potential energy reference at the center of the hemisphere. Initial state (at the top):

• Potential Energy $PE_{initial} = mgr$
• Kinetic Energy $KE_{initial} = 0$
• Total Initial Energy $E_{initial} = mgr$

At angular displacement θ, the block's vertical position relative to the center is $r \cos θ$. Final state (at angle θ):

• Potential Energy $PE_{final} = mgr \cos θ$
• Kinetic Energy $KE_{final} = (1/2)mv^2$ (where v is the speed of the block)
• Total Final Energy $E_{final} = mgr \cos θ + (1/2)mv^2$

By conservation of energy ($E_{initial} = E_{final}$): $mgr = mgr \cos θ + (1/2)mv^2$ $gr(1 - \cos θ) = (1/2)v^2$ $v^2 = 2gr(1 - \cos θ)$ (Equation 1)

2. Newton's Second Law for Circular Motion: At the point where the block is at angle θ, the forces acting on it are:

• Gravitational force mg acting vertically downwards.
• Normal force N acting radially outwards from the center of the sphere.

The component of gravity along the radius towards the center is $mg \cos θ$. The net radial force towards the center provides the centripetal force $mv^2/r$. So, applying Newton's second law in the radial direction: $mg \cos θ - N = mv^2/r$ (Equation 2)

3. Condition for losing contact: The block loses contact with the surface when the normal force N becomes zero. Setting N = 0 in Equation 2: $mg \cos θ = mv^2/r$ $v^2 = gr \cos θ$ (Equation 3)

4. Solve for θ: Equate the expressions for $v^2$ from Equation 1 and Equation 3: $2gr(1 - \cos θ) = gr \cos θ$ Divide both sides by gr (assuming g and r are non-zero): $2(1 - \cos θ) = \cos θ$ $2 - 2 \cos θ = \cos θ$ $2 = 3 \cos θ$ $\cos θ = 2/3$ $θ = \cos^{-1}(2/3)$

Part (b): Horizontal displacement of the block when it hits the floor.

After losing contact, the block undergoes projectile motion. The hemisphere is fixed on a horizontal floor, meaning the center of the hemisphere is at a height r above the floor.

1. State at the point of losing contact (start of projectile motion):

• Angular displacement $θ = \cos^{-1}(2/3)$.
• We have $\cos θ = 2/3$.
• $\sin θ = \sqrt{1 - \cos^2 θ} = \sqrt{1 - (2/3)^2} = \sqrt{1 - 4/9} = \sqrt{5/9} = \frac{\sqrt{5}}{3}$.
• Speed $v = \sqrt{gr \cos θ} = \sqrt{gr \cdot \frac{2}{3}} = \sqrt{\frac{2gr}{3}}$.
• Coordinates of the block (taking the center of the hemisphere as (0, r) and the floor as y=0):
  • Horizontal position from the vertical axis through the center: $x_c = r \sin θ = r \frac{\sqrt{5}}{3}$.
  • Vertical position (height from the floor): $y_c = r + r \cos θ = r + r \frac{2}{3} = \frac{5r}{3}$.
• Velocity components: The velocity vector is tangential to the hemisphere. The angle it makes with the horizontal is θ.
  • Horizontal velocity component: $v_x = v \sin θ = \sqrt{\frac{2gr}{3}} \cdot \frac{\sqrt{5}}{3} = \frac{\sqrt{10gr}}{3}$.
  • Vertical velocity component: $v_y = -v \cos θ = -\sqrt{\frac{2gr}{3}} \cdot \frac{2}{3} = -\frac{2}{3}\sqrt{\frac{2gr}{3}}$. (Negative sign indicates downwards direction).

2. Time of flight (t) during projectile motion: The block falls from $y_c = \frac{5r}{3}$ to $y_f = 0$ (the floor). Using the kinematic equation for vertical motion: $y_f = y_c + v_y t + \frac{1}{2}a_y t^2$ $0 = \frac{5r}{3} - \frac{2}{3}\sqrt{\frac{2gr}{3}} t - \frac{1}{2}gt^2$ Rearrange into a quadratic equation for t: $\frac{1}{2}gt^2 + \frac{2}{3}\sqrt{\frac{2gr}{3}} t - \frac{5r}{3} = 0$ Multiply by 6: $3gt^2 + 4\sqrt{\frac{2gr}{3}} t - 10r = 0$ Using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $t = \frac{-4\sqrt{\frac{2gr}{3}} \pm \sqrt{(4\sqrt{\frac{2gr}{3}})^2 - 4(3g)(-10r)}}{2(3g)}$ $t = \frac{-4\sqrt{\frac{2gr}{3}} \pm \sqrt{16 \cdot \frac{2gr}{3} + 120gr}}{6g}$ $t = \frac{-4\sqrt{\frac{2gr}{3}} \pm \sqrt{\frac{32gr}{3} + \frac{360gr}{3}}}{6g}$ $t = \frac{-4\sqrt{\frac{2gr}{3}} \pm \sqrt{\frac{392gr}{3}}}{6g}$ $t = \frac{-4\sqrt{\frac{2}{3}}\sqrt{gr} \pm \sqrt{\frac{392}{3}}\sqrt{gr}}{6g}$ $\sqrt{\frac{392}{3}} = \sqrt{\frac{196 \cdot 2}{3}} = \frac{14\sqrt{2}}{\sqrt{3}} = \frac{14\sqrt{6}}{3}$ $\sqrt{\frac{2}{3}} = \frac{\sqrt{6}}{3}$ $t = \frac{-4\frac{\sqrt{6}}{3}\sqrt{gr} \pm \frac{14\sqrt{6}}{3}\sqrt{gr}}{6g}$ Since t must be positive, we take the positive root: $t = \frac{(-4\frac{\sqrt{6}}{3} + \frac{14\sqrt{6}}{3})\sqrt{gr}}{6g}$ $t = \frac{\frac{10\sqrt{6}}{3}\sqrt{gr}}{6g} = \frac{10\sqrt{6}}{18} \sqrt{\frac{r}{g}} = \frac{5\sqrt{6}}{9}\sqrt{\frac{r}{g}}$

3. Total horizontal displacement (X_total): The horizontal displacement during projectile motion is $Δx = v_x t$. $Δx = \left(\frac{\sqrt{10gr}}{3}\right) \left(\frac{5\sqrt{6}}{9}\sqrt{\frac{r}{g}}\right)$ $Δx = \frac{5\sqrt{60}gr}{27g} = \frac{5\sqrt{4 \cdot 15}r}{27} = \frac{5 \cdot 2\sqrt{15}r}{27} = \frac{10\sqrt{15}r}{27}$

The total horizontal displacement from the center of the hemisphere when it hits the floor is the sum of the horizontal position at contact and the horizontal displacement during projectile motion: $X_{total} = x_c + Δx$ $X_{total} = r \frac{\sqrt{5}}{3} + \frac{10\sqrt{15}r}{27}$ $X_{total} = r \left( \frac{\sqrt{5}}{3} + \frac{10\sqrt{15}}{27} \right)$ To combine, find a common denominator (27): $X_{total} = r \left( \frac{9\sqrt{5}}{27} + \frac{10\sqrt{15}}{27} \right)$ $X_{total} = \frac{r}{27} (9\sqrt{5} + 10\sqrt{15})$

The final answer is $\boxed{\text{(a) } \cos^{-1}(2/3), \text{ (b) } \frac{r}{27}(9\sqrt{5} + 10\sqrt{15})}$.

The question has two parts, (a) and (b). For part (a), the answer is provided in the question itself as cos^{-1}(2/3). Our calculation confirms this. For part (b), the final expression is \frac{r}{27}(9\sqrt{5} + 10\sqrt{15}).

Explanation of the solution:

1. Part (a) - Loss of Contact:

   • Apply conservation of mechanical energy: Initial potential energy at the top (mgr) is converted into potential (mgr cos θ) and kinetic energy (1/2 mv^2) at an angle θ. This gives v^2 = 2gr(1 - cos θ).
   • Apply Newton's second law in the radial direction: The net force towards the center provides the centripetal force. This is mg cos θ - N = mv^2/r.
   • The condition for losing contact is N = 0. Substituting this into the force equation gives v^2 = gr cos θ.
   • Equating the two expressions for v^2 yields 2gr(1 - cos θ) = gr cos θ, which simplifies to cos θ = 2/3. Thus, θ = cos^{-1}(2/3).

2. Part (b) - Horizontal Displacement to Floor:

   • After losing contact, the block becomes a projectile.
   • Calculate the height (y_c = r + r cos θ) and horizontal position (x_c = r sin θ) of the block relative to the floor at the point of contact loss.
   • Calculate the horizontal (v_x = v sin θ) and vertical (v_y = -v cos θ) components of the velocity at the point of contact loss.
   • Use kinematic equations for projectile motion to find the time of flight (t) from y_c to 0 under gravity (g). This involves solving a quadratic equation for t.
   • Calculate the horizontal distance covered during this time (Δx = v_x t).
   • The total horizontal displacement is the sum of x_c and Δx.

Answer: (a) The angular displacement is \cos^{-1}(2/3). (b) The horizontal displacement of the block when it hits the floor is \frac{r}{27}(9\sqrt{5} + 10\sqrt{15}).