Answer:
[tex]-16.6 rad/s^2[/tex]
Explanation:
The torque exerted on a rigid body is related to the angular acceleration by the equation
[tex]\tau = I \alpha[/tex] (1)
where
[tex]\tau[/tex] is the torque
I is the moment of inertia of the body
[tex]\alpha[/tex] is the angular acceleration
Here we have a solid sphere: the moment of inertia of a sphere rotating about is centre is
[tex]I=\frac{2}{5}MR^2[/tex]
where
M = 240 g = 0.240 kg is the mass of the sphere
[tex]R=\frac{2.50}{2}=1.25 cm = 0.0125 m[/tex] is the radius of the sphere
Substituting,
[tex]I=\frac{2}{5}(0.240)(0.0125)^2=1.5\cdot 10^{-5} kg m^2[/tex]
The torque exerted on the sphere is
[tex]\tau = Fr[/tex]
where
F = -0.0200 N is the force of friction
r = 0.0125 m is the radius of the sphere
So
[tex]\tau=(-0.0200)(0.0125)=-2.5\cdot 10^{-4} Nm[/tex]
Substituting into (1), we find the angular acceleration:
[tex]\alpha = \frac{\tau}{I}=\frac{-2.5\cdot 10^{-4}}{1.5\cdot 10^{-5}}=-16.6 rad/s^2[/tex]
