A yo-yo has a rotational inertia of 790 g·cm2 and a mass of 95 g. Its axle radiu

Question

A yo-yo has a rotational inertia of 790 g·cm2 and a mass of 95 g. Its axle radius is 2.2 mm and its string is 100 cm long. The yo-yo is thrown so that its its initial speed down the string is 1.2 m/s.
(a) How long does the yo-yo take to reach the end of the string?
(b) As it reaches the end of the string, what is its total kinetic energy?
(c) As it reaches the end of the string, what is its linear speed?
(d) As it reaches the end of the string, what is its translational kinetic energy?
(e) As it reaches the end of the string, what is its angular speed?
(f) As it reaches the end of the string, what is its rotational kinetic energy?

Explanation / Answer

Rotational inertia I = 790 g-cm^2 Mass m = 95 g Radius r = 2.2 mm = 0.22 cm Length of the string h = 100 cm Initial speed v = 1.2 m/s At the bottom of the yo-yo's descent, the potential energy mgh will be entirely converted to translational kinetic energy ½mv² and rotational kinetic energy ½I². From conservation of energy

    E = Kt + Kr or
mgh = ½m²r² + ½I²
        = ½(mr² + I)² = [2mgh / (mr² + I)]         = [(2*95*980*100)/(95*0.22*0.22 + 790)         = 153.07 rad/s Linear speed v = r = 0.22 * 153.07 = 33.67 cm/s Rotational Kinetic energy Kr = [ I / (mr² + I) ]*E                                              = 9256126.99 J Translationa Kinetic energy Kt = [ mr² / (mr² + I)]*E                                                = 53873 J Total kinetic energy E = Kr + Kt = 9309999.99 J We use our projectile equation
v² = 2ay to find the translational acceleration. Once we have found acceleration a, we may solve your choice of the equations y = ½at² or a = v / t for t to answer (a).

We use our projectile equation
v² = 2ay to find the translational acceleration. Once we have found acceleration a, we may solve your choice of the equations y = ½at² or a = v / t for t to answer (a).

     

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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