A merry-go-round accelerates from rest to 0.75 rad/s in 33 s.

Physics

1 answer:

Aleks04 [339]2 years ago

6 0

Answer:

1309.1 Nm

Explanation:

Torque is given as a product of Moment of innertia and acceleration hence

T=Ia where T is torque and a is acceleration

To get acceleration, it is rate of change of speed per unit time hence $a=\frac {v_f-v_i}{t}$ where v and t represent velocity and time respectively while subscripts f and i represent final and initial respectively. Also, I is given by $0.5mr^{2}$ where m js mass and r is radius hence the net torque can now be written as

$T=0.5mr^{2}\times \frac {v_f-v_i}{t}$

By substituting the given figures then

$T=0.5\times 3.2\times 10^{4}\times 6^{2}\times \frac {0.75-0}{33}=1309.0909090867 Nm\approx 1309.1 Nm$

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3 years ago

You have a pendulum clock made from a uniform rod of mass M and length L pivoting around one end of the rod. Its frequency is 1

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The new oscillation frequency of the pendulum clock is 1.14 rad/s.

The given parameters;

Mass of the pendulum, = M
Length of the pendulum, = L
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$I_i = \frac{1}{3} ML^2$

The moment of inertia of the rod between the middle and the end is calculated as;

$I_f = \int\limits^L_{L/2} {r^2\frac{M}{L} } \, dr = \frac{M}{3L} [r^3]^L_{L/2} = \frac{M}{3L} [L^3 - \frac{L^3}{8} ] = \frac{M}{3L} [\frac{7L^3}{8} ]= \frac{7ML^2}{24}$

Apply the principle of conservation of angular momentum as shown below;

$I _i \omega _i = I _f \omega _f\\\\\frac{ML^2}{3} (1 \ rad/s)= \frac{7ML^2}{24} \times \omega _f\\\\\frac{24 \times ML^2}{3 \times 7 ML^2} (1 \ rad/s)= \omega _f\\\\1.14 \ rad/s = \omega _f$

Thus, the new oscillation frequency of the pendulum clock is 1.14 rad/s.

Learn more about moment of inertia of uniform rod here: brainly.com/question/15648129

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2 years ago

The radius of a planet is 2400 km, and the acceleration due to gravity at its surface is 3.6 m/s2.

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Answer:

$3.1\cdot10^{23}\:\mathrm{kg}$

Explanation:

We can use Newton's Universal Law of Gravitation to solve this problem:

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