yo-yo's inertia?

Question

A yo-yo has a rotational inertia of 950 g cm^2 and a mass of 120 g. Its axle radius is 3.2 mm, and its string is 120 cm long. The yo-yo rolls from rest down to the end of the sting.

(a) What is the magnitude of its translational acceleration?

(b) How long does it take to reach the end of the string?

As it reaches the end of the string, what are its (c) translational speed, (d) translational kinetic energy, (e) rotational kinetic energy, and (f) rotational speed?

What equations did you use to solve this, and can you explain your answers?

Answer 1

velocity of centre is half of velocity of the falling edge (due to the position of the pivot)

forces acting on the roating boady: string (N, up), mg down, the mg is applied to center of mass so it;s not used in angular accel directly. ma=mg-N, there is a realtionship b/w a (transl accel) and alfa (ang accel)
dw/dt=alfa, w=alfa*t,
decomposing into two mostions, rotational and translational: total velocity in center is half of edge (going down) velocity. but since we decomposed the motion for rotational only motion the edge velocity is = to center velocity.
v=w*r =w*r=alfa*t*r = a*t => a= alfa*r,
torque =N*r =I*alfa, N=I*alfa/r
from: ma=mg-N: ma =mg - I*alfa/r => alfa*I/r=mg- ma =m (g-alfa*r)=> alfa= mg /(I/r +m*r) =1.2N /(9.5x10^-5 kgm^2/3.2x10^-3 +0.12kg*3.2x10-3 )=39.9 rad/s^2 , a= alfa*r= 39.9 * 3.2x10-3 =0.128 m/s^2

b) s=a * t^2 / 2 => t= sqrt(2*s/a) =4.33s
c) v=a*t=0.128 m/s^2* 6.12 s =0.554 m/s
d) Ek_t= mv^2/2= 0.12kg * (0.554 m/s )^2 /2 =0.0184 J
e) Ek_rot=I*w^2 /2= I (v/r)^2 /2 = I (v_centre*2/r)^2 /2 =1.423J

check: Etot= mgh=1.44J = 1.423J + 0.0184 J =>OK

f) w= v/r=0.554/ 3.2x10^-3=173.1 rad/s

PS. I hope it's correct, do it your self and compare, i might be wrong
PPS. I have too much time on my hands. if you have questions, email

Answer 2

The equations are on your e book. practice the equations. The yo yo has means means. because it drops, that's coverted into rotational and translational means. Transl. means is a million/2 m v^2 Rot. means is a million/2 I omega^2 means means is m g h Pot. means = Transl. + Rot. Energies Sum of the moments = I alpha Sum of the forces = m a I aint gonna sove it for ya. you try a minimum of.

Answer 3

The equations are in your book. Prepare the equations.

The yo yo has potential energy. As it drops, this is coverted into rotational and translational energy.

Transl. energy is 1/2 m v^2
Rot. energy is 1/2 I omega^2
Potential energy is m g h


Pot. Energy = Transl. + Rot. Energies

Sum of the moments = I alpha
Sum of the forces = m a

I aint gonna sove it for ya. You try at least.