A platform rotates at an angular speed of 2.2 rad/s. A block rests on the platform a distance of 0.3m from the axis. Static friction coeffient= 0.5. Without any external torque acting on the system, the block is moved towards the axis. Ignore the moment of interia of the platform. Determine the smallest distance in cm from the axis the block can be moved and still remain in place as the platform rotates.
I know that the answer is between 17 and 23cm.
well, friction must equal centripetal force. As radius increases, centripetal force gets bigger, and friction cannot hold it. So your problem bothers me as it is written as if the moving inward makes the block more unstable, which is not true.
So where is the block unstable?
mu*mg=m*w^2 r
.5*9.8=2.2^2 *r
r=1.01 meter from the center. Inside that radius, it will stay, outside, it will fly off.
To determine the smallest distance in cm from the axis that the block can be moved and still remain in place as the platform rotates, we need to consider the balance between the centripetal force and the static friction force.
The centripetal force required to keep the block in place can be calculated using the formula:
Fc = m * ω^2 * r
where Fc is the centripetal force, m is the mass of the block, ω is the angular speed, and r is the distance of the block from the axis.
Since the mass of the block is not given in the problem statement, we can ignore it since it cancels out when comparing the forces.
Now, let's consider the static friction force. The maximum static friction force (Ff max) that can be exerted by the platform can be calculated using the formula:
Ff max = μ * N
where μ is the coefficient of static friction and N is the normal force.
To find the normal force N, we need to consider the vertical forces acting on the block. Since the block is at rest on the platform, the normal force is equal to the gravitational force acting on the block.
N = m * g
where g is the acceleration due to gravity.
Now we can substitute the values into the equations and solve for the smallest distance.
Fc = Ff max
m * ω^2 * r = μ * m * g
r = (μ * g) / (ω^2)
Given that μ = 0.5, g = 9.8 m/s^2, and ω = 2.2 rad/s, we can substitute these values into the equation to find the smallest distance:
r = (0.5 * 9.8) / (2.2^2)
r ≈ 17.6 cm
Therefore, the smallest distance in cm from the axis that the block can be moved and still remain in place as the platform rotates is approximately 17.6 cm.