Answer:
frictional force = 0.52 N
Explanation:
diameter of turn table (D1) = 30 cm = 0.3 m
mass of turn table (M1) = 1.2 kg
diameter of shaft (D2) = 1.2 cm = 0.012 m
mass of shaft (M2) = 450 g = 0.45 kg
time (t) = 15 seconds
acceleration due to gravity (g) = 9.8 m/s^{2}
radius of turn table (R1) = 0.3 / 2 = 0.15 m
radius of shaft (R2) = 0.012 / 2 = 0.006 m
total moment of inertia (I) = moment of inertia of turn table + moment of inertia of shaft
I = 0.5(M1)(R1)^{2} + O.5 (M2)(R2)^{2}
I = 0.5(1.2)(0.15)^{2} + O.5 (0.45)(0.006)^{2}
I = 0.0135 + 0.0000081 = 0.0135081
ω₁ = 33 rpm = 33 x \frac{2π}{60} = 3.5 rad/s
α = -ω₁/t = -3.5 / 15 = -0.23 rad/s^{2}
torque = I x α
torque = 0.0135081 x (-0.23) = - 0.00311 N.m
torque = frictional force x R2
- 0.00311 = frictional force x 0.006
frictional force = 0.52 N
The amount of friction force that the brake pad applied to the shaft is 0.52 N
To determine how much friction force does the brake pad applied to the shaft, we need to first know the moment of inertia of the solid turntable, followed by the angular acceleration of the turntable.
The moment of inertia can be computed by using the formula:
[tex]\mathbf{I = \dfrac{1}{2} MR^2}[/tex]
where;
[tex]\mathbf{I = \dfrac{1}{2} \times (1.2 \ kg) (\dfrac{0.3}{2})^2}[/tex]
I = 0.0135 kgm²
The angular acceleration of the solid turntable is also estimated by using the formula:
[tex]\mathbf{\alpha = \dfrac{\omega _f - \omega _i}{t}}[/tex]
where;
∴
[tex]\mathbf{\alpha = \dfrac{0 -33 rpm \times( \dfrac{2 \pi \ rad/s}{60 rpm})}{15\ s}}[/tex]
[tex]\mathbf{\alpha = -0.23 rad/s^2}[/tex]
Finally, determining the friction force by using the equation of torque;
[tex]\mathbf{\sum \tau = I \times \alpha}[/tex]
From dynamics of rotational motion;
[tex]\mathbf{r \times f= I \times \alpha}[/tex]
[tex]\mathbf{ f= \dfrac{I \times \alpha}{r}}[/tex]
where;
[tex]\mathbf{ f= \dfrac{0.0135 \ kg.m^2 \times 0.23 \ rad/s^2}{(\dfrac{1.2 }{2} \times 10^{-2} m)}}[/tex]
f ≅ 0.52 N
Therefore, we can conclude that the amount of friction force that the brake pad applied to the shaft is 0.52 N.
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