A platform rotates at an angular speed of 2.2 rad/s. A block rests on this platform a distance of 0.3m from the axis. (friction coefficient .5) Without any external torgue acting on the system, the block is moved toward the axis. Ignore the moment of inertia of the platform. Determine the smallest distance (in cm) from the axis the block can be relocated and still remain in place as the platform rotates
The moment of inertia of the platform with block will decrease as the block is pushed toward the center. This will cause the angular velocity of the platform to increase. When calculating the moment of inertia and angular velocity of block distance R, ignore the I of the platform (because they tell you to. (It seems an unrealistic assumption to me). The centripetal acceleration of the block will increase as it moves to the center. At some radial distance form the axis, a static friction coefficient of 0.5 will be necessary to keep the block from slipping. That will be the answer.
To determine the smallest distance from the axis the block can be relocated and still remain in place as the platform rotates, we need to consider the centripetal force acting on the block.
The centripetal force is given by the formula:
F = (m * ω^2 * r)
Where:
F is the centripetal force
m is the mass of the block
ω is the angular speed of the platform in radians per second
r is the distance of the block from the axis of rotation
In this case, we need to find the smallest value of r that still allows the block to remain in place, which means that the frictional force between the block and the platform must be equal to the centripetal force acting on the block.
The frictional force is given by the formula:
F_friction = μ * N
Where:
μ is the coefficient of friction
N is the normal force acting on the block
The normal force is equal to the weight of the block, which is given by:
N = m * g
Where:
g is the acceleration due to gravity
Since the block is stationary, the frictional force must equal the centripetal force:
F_friction = F
Therefore,
μ * N = m * ω^2 * r
Substituting the expression for N:
μ * (m * g) = m * ω^2 * r
Simplifying:
μ * g = ω^2 * r
Now, we can solve for r:
r = (μ * g) / ω^2
Given that the friction coefficient μ is 0.5, the acceleration due to gravity g is approximately 9.8 m/s^2, and the angular speed ω is 2.2 rad/s, we can calculate the value of r:
r = (0.5 * 9.8) / (2.2^2)
r ≈ 1.11 m
Finally, converting this distance from meters to centimeters:
r ≈ 111 cm
Therefore, the smallest distance from the axis the block can be relocated while still remaining in place as the platform rotates is approximately 111 cm.