The string in a yo-yo is wound around an axle of radius 0.549 cm. The yo-yo has

Question

The string in a yo-yo is wound around an axle of radius 0.549 cm. The yo-yo has both rotational and translational motion, like a rolling object, and has mass 0.296 kg and outer radius 1.98 cm. Starting from rest, it rotates and falls a distance of 1.09 m (the length of the string). Assume for simplicity that the yo-yo is a uniform circular disk and that the string is thin compared to the radius of the axle. (a) What is the speed of the yo-yo when it reaches the distance of 1.09 m? m/s (b) How long does it take to fall? [Hint: The translational and rotational kinetic energies are related, but the yo-yo is not rolling on its outer radius.] s

Explanation / Answer

a) Let's use conservation of energy m*g*h=.5*I*w^2+.5*m*v^2 since the yo-yo is spun by the torque on the axle, w=v/r, where r is the radius of the axle and I=m*R^2/2, where R is the outer radius so m*g*h=.5*.*R^2*v^2/(2*r)+.5*m*v^2 v=sqrt((2*g*h)/(R^2/(2*r)+1)) v=5.33 m/s b)T*r=I*alpha where T is the tension in the string and m*g-T=m*a again, the connection of the axle relates alpha=a/r where r is the radius of the axle so T=m*R^2*a/(2*r^2) T=m*(g-a) a=g/(R^2/(2*r^2)+1) a=1.31 m/s^2 using y(t)=-.5*a*t^2 when y(t)=-1.5 t=1.51 seconds

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