A puck of mass m = 1.90 kg slides in a circle of radius r = 16.0 cm on a frictionless table while attached to a hanging cylinder of mass M = 3.70 kg by a cord through a hole in the table. What speed keeps the cylinder at rest?
m1v^2/r = m2g so:
v = sqrt(m2gr/m1)
To find the speed that keeps the cylinder at rest, we can use the concept of centripetal force. The force that keeps the puck moving in a circle is provided by the tension in the string. This tension is also the force acting on the cylinder that opposes its weight and keeps it at rest.
The tension in the string can be calculated using the centripetal force equation:
T = m * v^2 / r
where T is the tension, m is the mass of the puck, v is the velocity of the puck, and r is the radius of the circle.
Since the cylinder is at rest, the tension must be equal to its weight (mg):
T = M * g
where M is the mass of the cylinder and g is the acceleration due to gravity.
Setting these two equations equal to each other, we have:
M * g = m * v^2 / r
We can rearrange this equation to solve for v:
v^2 = (M * g * r) / m
v = sqrt((M * g * r) / m)
Substituting the given values:
M = 3.70 kg
g = 9.8 m/s^2
r = 0.16 m
m = 1.90 kg
v = sqrt((3.70 kg * 9.8 m/s^2 * 0.16 m) / 1.90 kg)
v = sqrt(6.6256)
v ≈ 2.57 m/s
Therefore, a speed of approximately 2.57 m/s will keep the cylinder at rest.