Question

A 2.00-kg textbook rests on a frictionless, horizontal surface. A cord attached to the book passes over a pulley whose diameter is 0.150 m, to a hanging book with mass 3.00 kg. The system is released from rest, and the books are observed to move 1.20 m in 0.800 s. (a) What is the tension in each part of the cord? (b) What is the moment of inertia of the pulley about its rotation axis? Can you explain why the tensions are not the same even though it is the same cord.

Answer 1

Answer:

a) T = 7.5 N

   T' = 18.15 N

b)  I = 0.016 kgm²

Explanation:

Given that:

Mass of the textbook  m =  2 kg

Diameter of the pulley d = 0.150 m

Hanged mass m' = 3 kg

Displacement  s = 1.2 m

Time t = 0.800 s

According to kinematics equation

Displacement  s  can be derived from the second equation of motion:

$s = ut + \frac{1}{2} at ^2$

where u = 0

$s = \frac{1}{2} at ^2$

making acceleration a the subject of the formula; we have:

$a = \frac{2s}{t^2}$

$a = \frac{2*1.2}{0.8^2}$

$a = 3.75 m/s^2$

Now; taking into account of mass m;

The tension in the cord attached to the book on the horizontal surfacce can be calculated as:

T = ma

T = 2 × 3.75

T = 7.5 N

For the mass m; the tension is calculated as :

m'g - T' = m' a

T' = m'(g-a)

T' = 3 × (9.8 - 3.75)

T' = 18.15 N

b)

Considering the pulley:

$(T'-T) r = I\alpha$

where;

$\alpha = \frac{a}{r}$

Then

$(T'-T) r = I* \frac{a}{r}$

Then the moment of inertia I/ can be re-written as :

$I = (T'-T)\frac{r^2}{a}$

$I = (18.15-7.5)\frac{0.75^2}{3.75}$

I = 0.016 kgm²

Answer 2

Answer:

The answer is attached

Explanation: