Question

A particle rests on the top of the hemisphere of radius R. The small horizontal velocity that must be imparted to the particle if it is to leave the hemisphere without sliding down. is-

Answer 1

Answer:

• The horizontal velocity (v) imparted will be √ g R.

Given:

1. The radius of the hemisphere = R

Explanation:

$\rule{300}{1.5}$

# Refer the attachment for figure.

When the body will leave the top of the hemisphere, the Normal Reaction will be equal (or) less than the Centripetal Force.

Therefore,

⇒ N ≤ F_c

Where,

• N Denotes Normal Reaction.
• F_c Denotes Centripetal Force.

For the condition that the body just leaves the top, Both Normal reaction and centripetal force will be equal.

⇒ N = F_c

Substituting the values,

⇒ Mg = F_c

∵ [ N = M g ]

⇒ Mg = M v² / R

∵ [ F_c = M v² / R ]

Simplifying,

⇒ Mg = M v² / R

⇒ g = v² / R

⇒ g × R = v²

⇒ v² = g R

⇒ v = √ g R

⇒ v = √ g R

∴ The horizontal velocity (v) imparted will be √ g R.

$\rule{300}{1.5}$

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

$\bigstar \: \boxed{\sf{ \large{\red{N = mg}}}}$

• Just after leave the position

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

⏭ Calculation:

✏ Comparing both the conditions, we get

$\Rightarrow \sf \: \green{ \cancel{ \purple{m}}g = \dfrac{ \cancel{ \purple{m }}{v}^{2} }{R}} \\ \\ \Rightarrow \sf \: {v}^{2} = gR \\ \\ \Rightarrow \: \underline{ \boxed{ \bold{ \sf{ \orange{ \large{v = \sqrt{gR}}}}}}} \: \: \gray{\surd }$