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:
[tex]s = ut + \frac{1}{2} at ^2[/tex]
where u = 0
[tex]s = \frac{1}{2} at ^2[/tex]
making acceleration a the subject of the formula; we have:
[tex]a = \frac{2s}{t^2}[/tex]
[tex]a = \frac{2*1.2}{0.8^2}[/tex]
[tex]a = 3.75 m/s^2[/tex]
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:
[tex](T'-T) r = I\alpha[/tex]
where;
[tex]\alpha = \frac{a}{r}[/tex]
Then
[tex](T'-T) r = I* \frac{a}{r}[/tex]
Then the moment of inertia I/ can be re-written as :
[tex]I = (T'-T)\frac{r^2}{a}[/tex]
[tex]I = (18.15-7.5)\frac{0.75^2}{3.75}[/tex]
I = 0.016 kgm²
