Question

10. A yo-yo is made by wrapping a 2-m long string around a 500-g 10-cm diameter uniform disk. The string is fastened to the ceiling, and the yo-yo is released. What is the tension in the string as the yo-yo falls?
A.9.8N
B.1.6N
C.4.9N
D. 0.49N
E.3.3N

The answer is B. How do I get this answer? I'm having trouble figuring out why tension would be less than (mg). Thank you in advance.

Answer 1

Tension = T

force up = T
force down = m g
m g - T = m a

torque = I * angular acceleration
T * R = I alpha = I a/R
but I for disk = (1/2) m R^2
so
T R = (1/2) m R a
T = (1/2) m a interesting :)
so
m g - (1/2) m a = m a
g = (3/2) a
a = (2/3) g
so
T = (1/2) m (2/3) g = (1/3) m g
= (1/3)(0.5)(9.81)
= 1.63 N or B

Answer 2

T-tension, M –torque,
I = mR²/2 -moment of inertia of disc
ε-angular acceleration

T•R= M
I•ε =M
ε =a/R
I•a/R = T•R,
a=TR²/I= 2TR²/mR²=2T/m,
ma=mg-T
m•2T/m = mg-T,
T=mg/3 =0.5•9.8/3 = 1.63
ANS: B