Question

A hollow sphere has radius 6.4 m. what is the minimum velocity required by a motor
cyclist at bottom to complete the circle.
​

Answer 1

Answer:

17.7 m/s

Explanation:

At the top of the sphere, mv²/r = mg

⇒ v² = rg,    at this point,

Total energy = KE + PE

              = 1/2 (mv²) + (mg(2r))

              = 1/2 (mrg) + 2mgr

Total energy would never change, so at the bottom, where potential energy is 0,   let the velocity be u.

⇒ total energy = KE

⇒ 1/2 (mrg) + 2mgr = 1/2 mu²

⇒ 5mrg = mu²

⇒ √5rg = u

⇒ √(5 x 6.4 x 9.8) = u

⇒ 17.7 m/s = u

Answer 2

$\huge\rm{\underline{\underbrace{Question:}}}$

A hollow sphere has radius 6.4m. Minimum velocity required by a motor cyclist at bottom to complete the circle will be?

$\huge\rm{\underline{\underbrace{Solution:}}}$

$\implies\sf{\dfrac{mv^2_0}{R} = mg}$

$\implies\sf{\dfrac{mgR}{m} = v^2_0}$

$\implies\sf{v^2_0 = gR}$

We know that from conservation of energy:

$\implies\bf{K_1+U_1=K_2+U_2}$

Where,

• $\bf{K_1}$ = Initial kinetic energy

• $\bf{K_2}$ = Final kinetic energy

• $\bf{U_1}$ = Initial potential energy

• $\bf{U_2}$ = Final potential energy

Note:

• Initial potential energy= 0
• Height = 2R

$\implies\rm{\dfrac{1}{2} mv^2+0 = \dfrac{1}{2} mv^2_0+mg(2R)}$

$\implies\rm{v^2 = gR+4gR}$

$\implies\rm{v^2 = 5gR}$

$\implies\rm{v = \sqrt{5gr}}$

$\implies\rm{v = \sqrt{5(9.8)(6.4)}}$

$\implies\rm{v = \sqrt{313.6}}$

$\green{\boxed{\implies\rm{v = 17.7ms^{-1}}}}$