A yo-yo has a rotational inertia of 1140 g.cm^2 and a mass of 138 g. Its axle ra

Question

A yo-yo has a rotational inertia of 1140 g.cm^2 and a mass of 138 g. Its axle radius is 3.55 mm, and its string is 108 cm long. The yo-yo rolls from rest down to the end of the string. (a) What is the magnitude of its linear acceleration? (b) How long does it take to reach the end of the string? As it reaches the end of the string, what are Rs (c) linear speed, (d) translational kinetic energy, (e) rotational kinetic energy, and (f) angular speed? Number Units Number Units Number Units Number Units Number Units Number units

Explanation / Answer

Here, m = 138 g = 0.138 kg

So, weight = mg = 0.138 * 9.81 = 1.35 N

I = 1140 g-cm^2 = 11.4 x 10^-5 kg-m^2

r = 3.55x10^-3 m

Force acting on the rotating body -

String (N, up), mg down, the mg is applied on mass so it is not used in angular acceleration directly.

So,

ma=mg-N,

Again, there is a relationship between a (translational acceleration) and alfa (angular acceleration)
dw/dt=alfa, w=alfa*t,
decomposing into two positions, rotational and translational –

Total velocity is half of edge (going down) velocity. But since we decomposed the motion for rotational only motion the edge velocity, so we have -

v=w*r =w*r=alfa*t*r = a*t => a= alfa*r,
torque =N*r =I*alfa, N=I*alfa/r
(a) From: ma=mg-N: ma =mg - I*alfa/r

=> alfa*I/r=mg- ma =m (g-alfa*r)

=> alfa= mg /(I/r +m*r) =1.35N /(11.4x10^-5 kgm^2/3.55x10^-3 +0.138kg*3.55x10-3 )

= 1.35 / (0.03211 + 0.00049) = 41.41 rad/s^2 ,

a= alfa*r = 41.41 * 3.55x10-3 = 147.0 x 10^-3 = 0.147 m/s^2

(b) s=a * t^2 / 2 => t= sqrt(2*s/a) = sqrt (2*1.08 / 0.147) = 3.83 s

(c) v = a*t= 0.147 m/s^2* 3.83 s = 0.563 m/s

(d) Ek_t = mv^2/2= 0.138kg * (0.563 m/s )^2 /2 =0.0219 J

(e) Ek_rot= I*w^2 /2= I (v/r)^2 /2 = I (v_centre*2/r)^2 /2 = 11.4x10^-5 x (0.563/0.00355)^2 = 2.87 J

(f) w= v/r=0.563/ 3.55x10^-3 = 158.6 rad/s

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