Question

A 2.0 kg stone tied to the the end of an inextensible string is whirled around in a horizontal circle of radius 1.5 m at auniform angular speed
2π
rad/s Calculate the rotational kinetic energy of the stone.

220.7 J

88.8 J

47.4 J

246.7 J

Answer 1

Answer on Question #41464 – Physics – Mechanics

A 2.0 kg stone tied to the end of an inextensible string is whirled around in a horizontal circle of radius 1.5 m at a uniform angular speed $2\pi$ rad/s. Calculate the rotational kinetic energy of the stone.

220.7 J

88.8 J

47.4 J

246.7 J

Solution:

$$m = 2kg - \text{mass of the stone};$$ $$\omega = 2\pi \frac{\text{rad}}{s} - \text{angular speed};$$ $$R = 1.5m - \text{radius of the circle};$$ $$J = mR^2 - \text{moment of inertia for the stone (Point mass } m \text{ at a distance } R \text{ from the axis of rotation)}$$

Formula for the rotational kinetic energy:

$$E_k = \frac{\omega^2 J}{2} = \frac{(2\pi)^2 \cdot mR^2}{2} = 2\pi^2 mR^2 = 2 \cdot \left(3.14 \frac{\text{rad}}{s}\right)^2 \cdot 2kg \cdot 1.5m^2 = 59.2J$$

Answer: rotational kinetic energy of the stone is equal to 59.2J.

http://www.AssignmentExpert.com/