Respuesta :
Answer:
Explanation:
There is change in moment of inertia of turntable due to which its angular velocity changes . We shall apply law of conservation of angular momentum
initial moment of inertia = 1 / 2 M r² , M is mass and r is radius of table
I₁ = 1/2 x 2.1 x .10²
= .0105 kg m²
final moment of inertia
I₂ = I₁ + 2 x m r² , m is mass of fallen blocks on the table
= .0105 + 2 x .52 x .1²
= .0105 + .0104
= .0209 kg m²
initial angular velocity ω₁ = 2π n , n is revolution per second
= 2π x 70 / 60
= 2.3333 π rad /s
Final angular velocity ω₂ = ?
Applying law of conservation of angular momentum
I₁ ω₁ = I₂ ω₂
.0105 x 2.3333 π = .0209 x ω₂
ω₂ = 1.172 π
= .586 x 2 π rad / s
no of revolution per sec = .586
no of revolution per minute = .586 x 60
= 35.16 rpm .
The turntable's angular velocity, in rpm, just after this event is 35.14 rpm.
The given parameters;
- mass of the turntable, M = 2.1 kg
- diameter of the turn table, d = 20 cm
- radius of the turn table, R = 10 cm
- mass of each of the two blocks, m = 520 g = 0.52 kg
- initial angular speed of the turn table, ω₁ = 70 rpm
The initial angular speed in rad/s is calculated as follows;
[tex]70 \ \frac{rev}{\min} \times \frac{1\min}{60 \ s} \times \frac{2\pi \ rad}{1 \ rev} = 7.331 \ rad/s[/tex]
The moment of inertia of the turntable is calculated as;
[tex]I = \frac{1}{2} MR^2\\\\I _1 = \frac{1}{2} \times 2.1 \times 0.1^2\\\\I_1 = 0.0105 \ kgm/s[/tex]
The momentum of inertia of the blocks is calculated as
[tex]I = MR^2 \\\\I_2 = 2MR^2\\\\I_2 = 2 \times 0.52 \times (0.1)^2 \\\\I_2 = 0.0104 \ kgm/s[/tex]
The turntable's angular velocity, in rpm, just after this event is calculated as follows;
[tex]I_i \omega _i = I_f \omega _f\\\\\omega _f = \frac{I_i \omega _i }{I_f} \\\\\omega _f = \frac{0.0105 \times 7.331}{0.0105 + 0.0104} \\\\\omega_f = 3.68 \ rad/s[/tex]
[tex]\omega _f = 3.68 \ rad/s \ \times \ \frac{1 \ rev}{2\pi \ rad} \times \frac{60 \ s}{1 \min} = 35.14 \ rpm[/tex]
Thus, the turntable's angular velocity, in rpm, just after this event is 35.14 rpm.
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