Bowling Ball Down a Slick Lane

Three forces act on the bowling ball: the gravitational force $Mg$ acting downward at the center of mass, the normal force $N$ acting upward at the contact point $C$, and the kinetic friction force $f_k = \mu_k N = \mu_k Mg$ acting backward (opposite to the velocity) at the contact point $C$.

To find the net torque about point $P$ on the floor, we analyze each force:

Gravitational force: The weight $Mg$ acts at the center of mass, which is at height $R$ above the floor.

Since $Mg$ points vertically downward, the torque magnitude about $P$ is

where $r_\perp$ is the perpendicular distance from $P$ to the vertical line of action of $Mg$, which is the horizontal distance from $P$ to the center of mass.

Normal force: The normal force $N = Mg$ acts at the contact point $C$ and points vertically upward.

Its line of action is the same vertical line as the line of action of $Mg$, so it has the same moment arm $r_\perp$. Thus its torque magnitude about $P$ is

and it is opposite in direction to $\tau_{gravity}$.

Friction force: The friction force $f_k$ acts horizontally at the contact point $C$, which lies on the floor, and point $P$ is also on the floor. Therefore, the perpendicular distance from $P$ to the horizontal line of action of $f_k$ is $0$, giving

The normal-force torque cancels the gravitational-force torque, and friction contributes zero torque about $P$. Therefore,