A yo-yo has a rotational inertia of 8.50 x 10-5 kg•m/s2 and a mass of 135 g. It

Question

A yo-yo has a rotational inertia of 8.50 x 10-5 kg•m/s2 and a mass of 135 g. It has a central axis of radius 3.0 mm and a 110 cm long string. If the yo-yo rolls from rest down to the bottom of the string, assuming the string has zero thickness, find

(a) the acceleration of the yo-yo.
(b) the time taken to reach the bottom.
(c) the linear speed at the bottom of the string.
(d) the angular speed at the bottom of the string.
(e) the linear kinetic energy at the bottom.
(f) the angular kinetic energy at the bottom.
(g) the total energy at the top of the string.

Explanation / Answer

I = 8.50 x 10-5 kg•m/s2, m = 135 g, r = 3.0 mm, L = 110 cm,
mgr = I, angular acceleration = mgr/I = 46.7 rad/s2
(a) the acceleration of the yo-yo a = r = 0.140 m/s2
(b) the time taken to reach the bottom = t

L = at2/2, so t = (2L/a) = 3.96 s
(c) the linear speed at the bottom of the string v = at = 0.555 m/s
(d) the angular speed at the bottom of the string = v/r = 185 rad/s
(e) the linear kinetic energy at the bottom = mv2/2 = 0.0208 J
(f) the angular kinetic energy at the bottom = I2/2 = 1.45 J
(g) the total energy at the top of the string = 0.0208 + 1.45 = 1.47 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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