A horizontal disk of radius $R = 10 m$ rotates about its axis with constant angular velocity $\omega = 0.2 rad/s$. A motorcyclist rides along the edge of the disk with constant velocity $v = 36 km/h$ relative to the disk in the direction opposite to the direction of rotation of the disk. What should be the coefficient of friction between the tires of the motorcycle and the surface of the disk, so that such a movement is possible?
Decision:
For a motorcycle to move on a circle of radius $R$ with velocity $v$, a force directed toward the center of the disc must be applied to it
$F = \frac{mv^{2}}{R}$, (1)
Where $m$ is the mass of the motorcycle with the motorcyclist. This formula is valid only in an inertial reference frame, such as that associated with the earth. In this system, the velocity of the motorcycle $v$ is the sum of the velocity of the motorcycle relative to the disk $v_{0}$ and the velocity of the points of the edge of the disk relative to the ground $u=\omega R$:
$v=v_{0}-u=v_{0}- \omega R$. (2)
The minus sign occurs because the direction of rotation of the disk is opposite to the direction of motion of the motorcycle. The only force acting on the motorcycle with the motorcyclist in the horizontal direction perpendicular to the speed of the motorcycle is the resting frictional force between the tire surface and the disc. This force cannot exceed its maximum value
$F_{max}=\mu |\bar{N}|$, (3)
where $\bar{N}$ is the force of normal pressure from the disk side on the motorcycle c motorcyclist and $\mu$ is the coefficient of friction. There are only two forces acting on them in the vertical direction: $m \bar{g}$ and $\bar{N}$. Since the motorcycle is not moving in this direction, $m \bar{g} + \bar{N} = 0$, whence
$|\bar{N}|=mg$, (4)
Considering the equalities (1) - (4), we obtain
$F= \frac{m(v- \omega R)^{2}}{R} \leq F_{max} = \mu mg$. (5)
Hence
$\mu \geq \frac{(v-\omega R)^{2}}{Rg}=0.64$.
So, to keep the bike from slipping off the disk, the coefficient of friction must be at least 0.64.
Decision:
For a motorcycle to move on a circle of radius $R$ with velocity $v$, a force directed toward the center of the disc must be applied to it
$F = \frac{mv^{2}}{R}$, (1)
Where $m$ is the mass of the motorcycle with the motorcyclist. This formula is valid only in an inertial reference frame, such as that associated with the earth. In this system, the velocity of the motorcycle $v$ is the sum of the velocity of the motorcycle relative to the disk $v_{0}$ and the velocity of the points of the edge of the disk relative to the ground $u=\omega R$:
$v=v_{0}-u=v_{0}- \omega R$. (2)
The minus sign occurs because the direction of rotation of the disk is opposite to the direction of motion of the motorcycle. The only force acting on the motorcycle with the motorcyclist in the horizontal direction perpendicular to the speed of the motorcycle is the resting frictional force between the tire surface and the disc. This force cannot exceed its maximum value
$F_{max}=\mu |\bar{N}|$, (3)
where $\bar{N}$ is the force of normal pressure from the disk side on the motorcycle c motorcyclist and $\mu$ is the coefficient of friction. There are only two forces acting on them in the vertical direction: $m \bar{g}$ and $\bar{N}$. Since the motorcycle is not moving in this direction, $m \bar{g} + \bar{N} = 0$, whence
$|\bar{N}|=mg$, (4)
Considering the equalities (1) - (4), we obtain
$F= \frac{m(v- \omega R)^{2}}{R} \leq F_{max} = \mu mg$. (5)
Hence
$\mu \geq \frac{(v-\omega R)^{2}}{Rg}=0.64$.
So, to keep the bike from slipping off the disk, the coefficient of friction must be at least 0.64.