A small block on a Frictionless, horizontal surface has a mass of 0.025() kg. It

Question

A small block on a Frictionless, horizontal surface has a mass of 0.025() kg. It is attached to a massless cord passing through a hole in the surface (Fig. E10.42). The block is originally revolving at a distance of 0.300 m from the hole with an angular speed of 1.75 rad/s. The cord is then pulled from below. shortening the radius of the circle in which the block revolves to 0.150 in. Model the block as a particle. (a) Is the angular momentum of the block conserved? Why or why not? (b) What is the new angular speed? (c) Find the change in kinetic energy of the block. (d) How much work was done in pulling the cord?

Explanation / Answer

a) angular momentum of the block is conserved.

Because, though there is a force on the particle towards the center,

net torque on the particle is zero.

b)Apply, I2*w2 = I1*w1

m*r2^2*w2 = m*r1^2*w1

==> w2 = w1*(r1/r2)^2

= 1.75*(0.3/0.15)^2

= 7 rad/s

c) KI = 0.5*I1*w1^2

= 0.5*m*r1^2*w1^2

= 0.5*0.025*0.3^2*1.75^2

= 0.003445 J

KF = 0.5*I2*w2^2

= 0.5*m*r2^2*w1^2

= 0.5*0.025*0.15^2*7^2

= 0.013781 J

KF - KI = 0.013781 - 0.003445

= 0.010336 J

D) Workdone = change in kinetic enrgy

= 0.010336 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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