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3 years ago

7

A 0.18-kg turntable of radius 0.32 m spins about a vertical axis through its center. A constant rotational acceleration causes t

he turntable to accelerate from 0 to 24 revolutions per second in 8.0 s.Calculate the rotational acceleration.

1 answer:

borishaifa [10]3 years ago

4 0

Answer:

Angular acceleration will be $18.84rad/sec^2$

Explanation:

We have given that mass m = 0.18 kg

Radius r = 0.32 m

Initial angular velocity $\omega _i=0rev/sec$

And final angular velocity $\omega _f=24rev/sec$

Time is given as t = 8 sec

From equation of motion

We know that $\omega _f=\omega _i+\alpha t$

$24=0+\alpha \times 8$

$\alpha =3rev/sec^2=3\times 2\times \pi rad/sec^2=18.84rad/sec^2$

So angular acceleration will be $18.84rad/sec^2$

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A 5.93 kg ball is attached to the top of a vertical pole with a 2.35 m length of massless string. The ball is struck, causing it

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3 years ago

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2 years ago

Read 2 more answers

Two astronauts (each with mass 100 kg) are drifting together through space. They are connected to each other by a rope 5 m in le

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Answer:

1000 kgm²/s, 400 J

1000 kgm²/s, 1000 J

600 J

Explanation:

m = Mass of astronauts = 100 kg

d = Diameter

r = Radius = $\frac{d}{2}$

v = Velocity of astronauts = 2 m/s

Angular momentum of the system is given by

$L=mvr+mvr\\\Rightarrow L=2mvr\\\Rightarrow L=2\times 100\times 2\times 2.5\\\Rightarrow L=1000\ kgm^2/s$

The angular momentum of the system is 1000 kgm²/s

Rotational energy is given by

$K=I\omega^2\\\Rightarrow K=\frac{1}{2}(mr^2)\left(\frac{v}{r}\right)^2\\\Rightarrow K=mv^2\\\Rightarrow K=100\times 2^2\\\Rightarrow K=400\ J$

The rotational energy of the system is 400 J

There no external toque present so the initial and final angular momentum will be equal to the initial angular momentum 1000 kgm²/s

$L_i=L_f\\\Rightarrow 2mv_ir_i=2mv_fr_f\\\Rightarrow v_f=\frac{v_ir_i}{r_f}\\\Rightarrow v_f=\frac{2\times 2.5}{0.5}\\\Rightarrow v_f=10\ m/s$

Energy

$E_2=mv_f^2\\\Rightarrow E_2=100\times 10\\\Rightarrow E_2=1000\ J$

The new energy will be 1000 J

Work done will be the change in the kinetic energy

$W=E_2-E\\\Rightarrow W=1000-400\\\Rightarrow W=600\ J$

The work done is 600 J

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3 years ago