A small block on a frictionless, horizontal surface has a mass of 0.0245 kg. It

Question

A small block on a frictionless, horizontal surface has a mass of 0.0245 kg. It is attached to a massless cord passing through a hole in the surface (see figure below). The block is originally revolving at a distance of 0.350 m from the hole with an angular speed of 1.90 rad/s. The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.175 m. Model the block as a particle.

A small block on a frictionless, horizontal surface has a mass of 0.0245 kg. It is attached to a massless cord passing through a hole in the surface (see figure below). The block is originally revolving at a distance of 0.350 m from the hole with an angular speed of 1.90 rad/s. The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.175 m. Model the block as a particle. Is angular momentum of the block conserved? Yes No Why or why not? What is the new angular speed? rad/s Find the change in kinetic energy of the block. L How much work was done in pulling the cord? J

Explanation / Answer

Yes the angular momentum would be conserved as there is no external torque being applied to change it.

angular momentum = moment of inertia x angular velocity

and moment of inertia = mass x distance^2 (since it is a point-mass)

hence you can work out the moment of inertia at the two distances then equate the two equations for angular momentum to find the new angular velocity.

0.025 x 0.3^2 x 1.75 = 0.025 x 0.15^2 x new angular velocity

the kinetic energy can be found with the equation

KE = 1/2 x moment of inertia x angular velocity^2

from which it should be easy to compute the difference

unless I am missing something I think the work done should be equal to the change in kinetic energy

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.