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.

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

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

To determine the tension in the string as the yo-yo falls, we need to consider the forces acting on the yo-yo.

The forces acting on the yo-yo are its weight (mg) and the tension in the string. When the yo-yo is released and falls, the tension in the string will counteract the yo-yo's weight to keep it in motion.

First, let's calculate the yo-yo's weight. The weight of an object is given by the formula: weight = mass x gravitational acceleration (mg).

Given that the mass of the yo-yo is 500 grams (0.5 kg) and the acceleration due to gravity is 9.8 m/s^2, the weight can be calculated as follows:
weight = 0.5 kg x 9.8 m/s^2 = 4.9 N

Now, let's consider the tension in the string. As the yo-yo falls, the tension will be exerted upwards to counteract its weight. The tension is also equal to the centripetal force required to keep the yo-yo moving in a circular path.

To determine the tension, we will use the formula for centripetal force:
centripetal force = mass x velocity^2 / radius

The mass of the yo-yo is 0.5 kg, and the radius of the disk is half its diameter, which is 10 cm (0.1 m). The velocity of the yo-yo can be calculated using the formula for the speed of an object in circular motion:
velocity = 2πr / T

The length of the string wrapped around the disk is given as 2 meters. Therefore, the circumference of the disk is 2πr, where r is the radius of the disk.

Using the formula for the circumference, we can calculate the radius:
2πr = 2 m
r = 1 m / π

Now, we can calculate the velocity:
velocity = 2π(1 m / π) / T = 2 m / T

Using this velocity in the formula for centripetal force, we have:
centripetal force = 0.5 kg x (2 m / T)^2 / 0.1 m = 10 kg m/s^2 / T^2

Since the tension is equal to the centripetal force, we can equate the two:
tension = 10 kg m/s^2 / T^2

Now, we can solve for T by setting the tension equal to the weight of the yo-yo:
T = sqrt(10 kg m/s^2 / 4.9 N) ≈ 1.6 N

Therefore, the tension in the string is approximately 1.6 N.

Hence, the correct answer is B.