Question

A uniform sphere of radius= R/16 starts rolling down without slipping from the top of another sphere of radius R=1m.Find the angular velocity of the sphere on rad^-1,after it leaves the surface of larger sphere

Answer 1

Answer of question #77153-Physics-Mechanics- Relativity

A uniform sphere of radius $= R / 16$ starts rolling down without slipping from the top of another sphere of radius $R = 1m$ . Find the angular velocity of the sphere on rad^1, after it leaves the surface of larger sphere

Input Data:

Radius: $R = 1$ m

Acceleration of gravity: $g = 9.81 \, \text{m/s}^2$

Solution:

Suppose that the loss of height is $H$ , then rolling speed of a small sphere:

$$v = \sqrt {2 * g * H};$$

The ball will detach from the sphere at the moment when the force of pressing is equal to the centrifugal force.

Centrifugal force: $F_{c} = \frac{mV^{2}}{R} = \frac{2*g*H*m}{R}$ ;

Force of pressing: $F_{\text{pressing}} = m * g * \cos(\varphi)$ ;

$$\cos (\varphi) = \frac {R - H}{R};$$

The condition for the separation of the ball:

$$F _ {c} = F _ {\text {p r e s s i n g}};$$

Simplifying the equation, we get:

$$2 H = R - H;$$ $$\mathrm {H} = \frac {R}{3};$$

Hence, we get the speed: $\nu = \sqrt{2 * g * R / 3}$ ;

Angular velocity: $\Omega = \frac{V}{R} = \sqrt{\frac{2g}{3R}}$ ;

Answer:

Angular velocity is 2.56 rad/s

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