Answer:
a) T = 7.5 N
T' = 18.15 N
b) I = 0.016 kgm²
Step-by-step 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 + (1)/(2) at ^2](https://img.qammunity.org/2021/formulas/physics/college/2km3bkxe24abj4jdcjokaf1u338n0uzhuj.png)
where u = 0
![s = (1)/(2) at ^2](https://img.qammunity.org/2021/formulas/physics/college/4xhxi4secz1kd5l847s89op3v7i07sel8w.png)
making acceleration a the subject of the formula; we have:
![a = (2s)/(t^2)](https://img.qammunity.org/2021/formulas/physics/college/p3x389rqnwkhyo4biktugu32g2vzm3ooyp.png)
![a = (2*1.2)/(0.8^2)](https://img.qammunity.org/2021/formulas/physics/college/xzagigzl2zhus0e7pz9ruhcg1c72fpn393.png)
![a = 3.75 m/s^2](https://img.qammunity.org/2021/formulas/physics/college/bd05j865wid3yesb7hxvhvs2fgi4u1bxvj.png)
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](https://img.qammunity.org/2021/formulas/physics/college/92vfkct00f74serzms8qw9k34o2xooovfk.png)
where;
![\alpha = (a)/(r)](https://img.qammunity.org/2021/formulas/physics/college/k1fzhixslodcnmx2tne3kgk18umtwlbf0s.png)
Then
![(T'-T) r = I* (a)/(r)](https://img.qammunity.org/2021/formulas/physics/college/7f1k30981mszglp6pl53b0ybddsa5ehmsy.png)
Then the moment of inertia I/ can be re-written as :
![I = (T'-T)(r^2)/(a)](https://img.qammunity.org/2021/formulas/physics/college/8gepma88e4io8zoup61fag3s1tokvjgsr5.png)
![I = (18.15-7.5)(0.75^2)/(3.75)](https://img.qammunity.org/2021/formulas/physics/college/wnt47ky0ynpvhrjqe5s2oyefg74jv7yaoc.png)
I = 0.016 kgm²