A small block on a frictionless horizontal surface has a mass of 2.50×10−2 . It is attached to a massless cord passing through a hole in the surface. (See the figure below .) The block is originally revolving at a distance of 0.300 from the hole with an angular speed of 1.75 . The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.150 . You may treat the block as a particle.
There is no question here. Perhaps they want to know what happens to the angular speed of revolution. It speeds up by a factor of 4, to keep the angular momentum constant. The mass is not important; it cancels out of the angular momentum conservation equation.
Is the 1.75 in radians per second? If so, the answer is 7.5 radians per second.
To solve this problem, we need to apply the principle of conservation of angular momentum. The formula for angular momentum is given by L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.
In the initial state, when the block is revolving at a distance of 0.300 m, the angular momentum is L1 = I1ω1. Here, I1 is the moment of inertia of the block at that distance, and ω1 is the initial angular speed.
In the final state, when the radius is shortened to 0.150 m, the angular momentum is L2 = I2ω2. Here, I2 is the moment of inertia of the block at the shortened distance, and ω2 is the final angular speed.
Since there is no external torque acting on the system, the angular momentum is conserved. Therefore, L1 = L2. We can set up the equation as follows:
I1ω1 = I2ω2
To find the moment of inertia of the block at each distance, we can use the formula for the moment of inertia of a point mass rotating about an axis perpendicular to it: I = mr^2, where m is the mass of the block and r is the distance from the axis of rotation.
Initially, the mass of the block is given as 2.50×10^(-2) kg, and the distance is 0.300 m. So, the moment of inertia in the initial state, I1 = (2.50×10^(-2) kg)(0.300 m)^2.
After shortening the radius to 0.150 m, the moment of inertia becomes I2 = (2.50×10^(-2) kg)(0.150 m)^2.
Now we can substitute these values back into the equation:
(2.50×10^(-2) kg)(0.300 m)^2 ω1 = (2.50×10^(-2) kg)(0.150 m)^2 ω2
Simplify the equation by canceling out the mass term:
(0.300 m)^2 ω1 = (0.150 m)^2 ω2
Now we can plug in the given values:
(0.300 m)^2 (1.75 rad/s) = (0.150 m)^2 ω2
Solve for ω2:
ω2 = [(0.300 m)^2 (1.75 rad/s)] / (0.150 m)^2
Evaluate the expression:
ω2 = (0.300 m)^2 (1.75 rad/s) / (0.150 m)^2
ω2 ≈ 4.17 rad/s
Therefore, the final angular speed of the block when the radius is shortened to 0.150 m is approximately 4.17 rad/s.