Question

A block is placed at the top of a smooth hemisphere of radius R. Now the hemisphere is given a horizontal acceleration $\sf a_0$. Find the velocity of the block relative to the hemisphere as a function of $\sf theta$as it slides down.​

Answer 1

Given : A block is placed at the top of a hemisphere with radius R.

• $\sf W_g$ = mg × y = mgR(1-cosθ)

• $\sf W_{pseudo}$ = m$\sf a_0$ × x = m$\sf a_0$ × R sin θ

• $\sf W_N$ = 0, because the velocity at the highest point is 0.

• $\sf W_g$ + $\sf W_{pseudo}$ + $\sf W_N$ = ∆K.E.

• mgR(1-cosθ) + m$\sf a_0$ × R sin θ + 0 = ½mv²

• v² = $\frac{2m[gR(1-cosθ) + [tex]\sf a_0$ [× R sin θ]}{m}[/tex]

• v = √2R[$\sf a_0$ sinθ + g(1 - cosθ)]

$\bold{Hope\;it \; helps\;!}$

Answer 2

Solution :

⏭ Given:

✏ Radius of hemisphere = R

⏭ To Find:

✏ The small horizontal velocity that must be imparted to the particle if it is to leave the hemisphere without sliding down.

⏭ Concept:

✏ This question is completely based on concept of Force equilibrium.

For rest position

$\bigstar \: \boxed{\sf{ \large{\red{N = mg}}}}</p><p>$

Just after leave the position

$\bigstar \: \boxed{ \sf{ \blue{ \large{N = \dfrac{m {v}^{2} }{R} }}}}$

⏭ Calculation:

✏ Comparing both the conditions, we get

$\begin{gathered} \Rightarrow \sf \: \green{ \cancel{ \purple{m}}g = \dfrac{ \cancel{ \purple{m }}{v}^{2} }{R}} \\ \\ \Rightarrow \sf \: {v}^{2} = gR \\ \\ \Rightarrow \: \underline{ \boxed{ \bold{ \sf{ \orange{ \large{v = \sqrt{gR}}}}}}} \: \: \gray{\surd }\end{gathered} </p><p>$