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A 1.9 kg , 20-cm-diameter turntable rotates at 150 rpm on frictionless bearings. Two 520 g blocks fall from above, hit the turntable simultaneously at opposite ends of a diameter, and stick. What is the turntable's angular velocity, inrpm, just after this event?

Respuesta :

Answer: 73.08 rpm

Explanation:

Given

Mass of the table, m(t) = 1.9 kg

Radius of table, r = 20/2 cm = 10 cm = 0.1 m

Mass of the block, m(b) = 500 g = 0.5 kg

Speed of rotation, w(i) = 150 rpm

Using conservation of angular momentum, where, initial angular momentum is equal to final angular momentum

L(i) = L(f)

I(i)w(i) = I(f)w(f)

w(f) = I(i)w(i) / I(f)

also, moment of inertia a disk or cylinder is I = 1/2mr²

Thus, I(i) = 1/2 mr²

I(i) = 1/2 * 1.9 * 0.1²

I(i) = 0.0095 kgm²

I(f) = Σm(i)r(i)² = I(i) + I(block1) + I(block2)

I(f) = I(i) + I(a) + I(b)

I(f) = 0.0095 + m(a)r(a)² + m(b)r(b)²

Remember, r(a) = r(b) = R

m(a) = m(b) = M, so that

I(f) = 0.0095 + MR² + MR²

I(f) = 0.0095 + 2MR²

I(f) = 0.0095 + 2 * 0.5 * 0.1²

I(f) = 0.0095 + 0.01

I(f) = 0.0195 kgm²

Substituting for values in the first equation

w(f) = I(i)w(i) / I(f)

w(f) = (0.0095 * 150) / 0.0195

w(f) = 1.425 / 0.0195

w(f) = 73.08 rpm

Thus, the turntable's angular velocity is 73.08 rpm

onyebuchinnaji

The turntable's angular velocity just after this event is 71.5 rpm.

The given parameters;

  • mass of table, m₁ = 1.9 kg
  • diameter of the table, d = 20 cm = 0.2 m
  • angular speed, ω₁ = 150 rpm
  • mass of the blocks, m₂ = 520 g = 0.52 kg

The final angular speed of the turntable is determined by applying the principle of conservation of angular momentum;

[tex]L_i = L_f\\\\I_i\omega_i = I_fw_f[/tex]

where;

  • [tex]\omega _i[/tex] is the initial angular speed

[tex]\omega _i = 150 \ \times \frac{rev}{\min} \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1\min}{60 \ s} = 15.71 \ rad/s[/tex]

[tex]\\\\(\frac{1}{2} MR^2)\omega _i = \omega _f (\frac{1}{2} MR^2 \ + \ m_1 R^2 \ + m_2R^2)\\\\(0.5 \times 1.9 \times 0.1^2)\times 15.71 = \omega _f[0.5\times 1.9( 0.1)^2\ + \ 0.52(0.1)^2 \ + \ 0.52( 0.1)^2]\\\\0.149 = \omega_f (0.0199)\\\\\omega_f = \frac{0.149}{0.0199} \\\\\omega_f = 7.487 \ rad/s\\\\\omega_f = 7.487 \ rad/s \ \times \ \frac{1 \ rev}{2\pi \ rad} \times \frac{60 \ s}{1 \min} \\\\\omega_f = 71.5 \ rpm[/tex]

Thus, the turntable's angular velocity just after this event is 71.5 rpm.

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