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Angelina_Jolie [31]
3 years ago
13

A person of mass 75 kg stands at the center of a rotating merry-go-round platform of radius 3.3 m and moment of inertia

Physics
1 answer:
vivado [14]3 years ago
7 0

Answer:

(a) 0.426 rad/s

(b) Before 296.225 J and After 148.5 J

Explanation:

At the center of merry-go-round, the person's weight is 75 Kg with moment of inertia of 820 kgm2 and the initial angular velocity is 0.85 rad/s

At the edge, the new angular velocity gained depends on the new moment of inertia

Here, the final moment of inertia is given by Initial moment of inertia + MR^{2} where M is the mass and R is the radius of merry-go-round

Final moment of inertia=820+(75*3.3^{2})=1636.75

From the law of conservation of angular momentum as torque is zero

I_i\times\omega_1=I_f\times\omega_2  and making \omega_2 the subject then

\omega_2=\frac {I_i\times\omega}{I_f}

where I_i and I_f are initial and final moment of inertia, \omega_1 and \omega_2 are initial and final angular velocity.

Substituting the provided values

\omega_2=\frac {0.85\times 820}{1636.75}=0.425843898\approx 0.426 rad/s

(b)

Initial rotational kinetic energy is given by

0.5I_i\times \omega_1^{2}

KE_i=0.5\times 820\times 0.85^{2}=296.225 J

The final rotational kinetic energy is given by

0.5I_f\times \omega_2^{2}

KE_i=0.5\times 1636.75\times 0.426^{2}\approx 148.5 J

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A wheel initially spinning at wo = 50.0 rad/s comes to a halt in 20.0 seconds. Determine the constant angular acceleration and t
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Answer:

(I). The angular acceleration and number of revolution are -2.5 rad/s² and 500 rad.

(II). The torque is 84.87 N-m.

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Initial spinning = 50.0 rad/s

Time = 20.0

Distance = 2.5 m

Mass of pole = 4 kg

Angle = 60°

We need to calculate the angular acceleration

Using formula of angular velocity

\omega=-\alpha t

\alpha=-\dfrac{\omega}{t}

\alpha=-\dfrac{50.0}{20.0}

\alpha=-2.5\ rad/s^2

The angular acceleration is -2.5 rad/s²

We need to calculate the number of revolution

Using angular equation of motion

\theta=\omega_{0}t+\dfrac{1}{2}\alpha t

Put the value into the formula

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\theta=500\ rad

The number of revolution is 500 rad.

(II). We need to calculate the torque

Using formula of torque

\vec{\tau}=\vec{r}\times\vec{f}

\tau=r\times f\sin\theta

Put the value into the formula

\tau=2.5\times4\times 9.8\sin60

\tau=84.87\ N-m

Hence, (I). The angular acceleration and number of revolution are -2.5 rad/s² and 500 rad.

(II). The torque is 84.87 N-m.

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\\ \sf\longmapsto x-20=30

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Option B

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