A small metal cylinder rests on a circular turntable that is rotating at a constant speed.
The small metal cylinder has a mass of 0.20kg, the coefficient of static friction between the cylinder and the turntable is 0.080, and the cylinder is located 0.15m from the center of the turntable.
Take the magnitude of the acceleration due to gravity to be 9.81 m/s^2.
What is the maximum speed v_max that the cylinder can move along its circular path without slipping off the turntable?
The answer is 0.3429.
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Well, it sounds like this metal cylinder is quite the daredevil, trying to hold on for dear life on that spinning turntable! Let's see if we can calculate the maximum speed it can reach without slipping off.
To figure this out, we'll need to consider the forces acting on the cylinder. The main force at play here is friction. The maximum static friction force is given by the equation F_static = μ_s * N, where μ_s is the coefficient of static friction and N is the normal force.
In this case, the normal force is equal to the gravitational force acting on the cylinder, which can be calculated as N = m * g, where m is the mass of the cylinder and g is the acceleration due to gravity.
Now, the friction force can be written as F_fric = m * g * μ_s.
Since the cylinder is moving in a circle, the centripetal force required to keep it on the circular path is given by F_centripetal = m * a_c, where a_c is the centripetal acceleration. The centripetal acceleration can be calculated as a_c = v^2 / r, where v is the velocity and r is the radius of the circular path.
In order for the cylinder not to slip, the friction force must provide the necessary centripetal force. So we can equate the two forces: m * g * μ_s = m * a_c.
Substituting the values given in the problem, we get:
0.2 * 9.81 * 0.080 = 0.2 * (v^2 / 0.15).
After doing some math, we find that the maximum speed v_max is approximately 1.66 m/s.
So, the cylinder can reach a maximum speed of 1.66 m/s along its circular path without slipping off the turntable. Just be careful not to make any sudden stops or sharp turns!
To find the maximum speed v_max that the cylinder can move along its circular path without slipping off the turntable, we need to consider the balance of forces acting on the cylinder.
The force that keeps the cylinder from slipping off the turntable is the static friction force. The maximum static friction force (F_max) can be calculated using the formula:
F_max = coefficient of static friction × normal force
The normal force (N) acting on the cylinder is equal to the weight of the cylinder, which is:
N = mass × acceleration due to gravity
Substituting the given values:
N = 0.20kg × 9.81 m/s^2
Next, we can calculate the maximum static friction force:
F_max = 0.080 × (0.20kg × 9.81 m/s^2)
Now, the maximum static friction force provides the centripetal force required to keep the cylinder moving in a circular path, which is equal to:
F_max = mass × (v_max)^2 / radius
Where mass is the mass of the cylinder and radius is the distance of the cylinder from the center of the turntable.
Rearranging the equation, we can solve for v_max:
(v_max)^2 = (F_max × radius) / mass
Taking the square root of both sides, we get:
v_max = √((F_max × radius) / mass)
Now, plugging in the known values:
v_max = √((F_max × 0.15m) / 0.20kg)
Evaluating the expression will give you the maximum speed v_max that the cylinder can move along its circular path without slipping off the turntable.
Refer to the following figure that shows the free-body diagram:
http://img340.imageshack.us/img340/4175/tiffanyc.jpg
F is the centrifugal force pushing the cylinder outwards, where
F=mv²/r
and v the tangential speed.
μN=μmg is the frictional force resisting the movement.
By equating
mv²/r = μmg
and solving for v, you will find
v=√(μgr)
I believe it evaluates to 0.3 m/s.
Check my thinking and calculations.