A merry-go-round with a radius of R = 1.65 m and moment of inertia I = 214 kg-m2

Question

A merry-go-round with a radius of R = 1.65 m and moment of inertia I = 214 kg-m2 is spinning with an initial angular speed of ? = 1.47 rad/s, in the counter-clockwise direction when viewed from above. A person with mass m = 66 kg and velocity v = 4.8 runs on a path tangent to the merry-go-round. Once at the merry-go-round the person jumps on and holds on to the rim of the merry-go-round.
1)What is the magnitude of the initial angular momentum of the merry-go-round?
2)What is the magnitude of the initial angular momentum of the person 2 meters before she jumps on the merry-go-round?
3)What is the magnitude of the initial angular momentum of the person just before she jumps on to the merry-go-round?
4)What is the angular speed of the merry-go-round after the person jumps on?

Explanation / Answer

Given that

R = 1.65 m

I = 214 kg-m2

= 1.47 rad/s

m = 66 kg

v = 4.8 m/s

a) Angular momentum of the merry go-round

               Lm = I

                  = 214 x 1.47

                  = 314.58 kg-m2/s

b) Given that

   here r = 2 m

So   L = m v r

            = 66 x 4.8 x 2

            = 633.6 kg-m2 /s

(c)

          Lp = m v R

            = 66 x 4.8 x 1.65

            = 522.72 kg-m2 /s

d) f = [Lm + Lp] / Ic

         = [Lm + Lp] / I + Ip

         = [Lm + Lp] / I + m R2

         = [314.58 + 522.72] / 214 + 66 x 1.652

         = 2.126 rad / s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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