Question

eyPLUS Mday, F 10e Its axie radius is 3.59 mm, and its string is 121 cm long. The yoryo rolls from rest down to the end of the the end of the string? As it reaches the end of the string, what are its (c) (d) translational kinetic energy, (e) rotational kinetic energy, and () angular speed? (b) N Units Il s Units TI m/s

Answer 1

Given : I = 818 g-cm²; m= 130 g ; r = 3.59 mm ; l= 121 cm

Solution :

(a)

Magnitude of linear acceleration is given by :

a = g /[ 1+(I/Mr²)]

Where g = 980 cm/s²

a = (980 cm/s²)÷[1+(818 g-cm²/130g *(0.359 cm)²)]

a = 19.67 cm/s² or 0.1967 m/s²

(b)

Applying equation of motion as :

y = v_it+(1/2)at²

-121 cm =0+ (1/2)(-19.67 cm/s²)t²

t²= 12.30

or t = 3.51 s

(c)

v = v₀+ at

Now putting values to get linear speed

v = 0-(19.67 cm/s²)*(3.51 s)

= 69.04 cm/s or 0.69 m/s

(d)

translational kinetic energy : KE = (1/2)mv²

KE = (1/2)*(0.130 kg)*(0.69m/s)²

= 0.031 J

(f)

Angular speed ( $\omega$ )= v/R = 69.04/0.359=192.31 rad/s

(e)

Rotational KE = (1/2)I $\omega$ ²

= (1/2)( 81.8 kg.m²)(192.31)²

= 1.5 J