A 7.27- kilogram bowling ball is rotating at 15.2 revolutions per second. A bowling ball had a diameter of 21.6 centimeters. Tom touches the spinning ball with his finger to stop the ball from rotating. How much torque is required to stop the rotation of the ball on 0.525 seconds?

Respuesta :

angular speed of the ball = 15.2 rev/s

diameter = 21.6 cm

radius = 10.8 cm

mass = 7.27 kg

Moment of inertia of ball is given as

[tex]I = \frac{2}{5}mR^2[/tex]

[tex]I = \frac{2}{5}\time 7.27\times (0.108)^2[/tex]

[tex]I = 0.034 kg m^2[/tex]

now angular acceleration is defined as rate of change in angular velocity

[tex]\alpha = \frac{\omega_f - \omega_i}{t}[/tex]

[tex]\alpha = \frac{2\pi (15.2) - 0}{0.525}[/tex]

[tex]\alpha = 182 rad/s[/tex]

now we know that

[tex]\tau = I \alpha[/tex]

[tex]\tau = 0.034 \times 182 = 6.18 Nm[/tex]

so it requires a torque of 6.18 Nm

ACCESS MORE
ACCESS MORE
ACCESS MORE
ACCESS MORE