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lakkis
1 day ago
12

A light-weight potter’s wheel, having a moment of inertia of 24 kg m2 is spinning freely at 40.0 rpm. The potter drops a small b

ut dense lump of clay onto the wheel, where it sticks a distance 1.2 m from the rotational axis. If the subsequent angular speed of the wheel and clay is 32 rpm, what is the mass of the clay?
Physics
1 answer:
Maru [2.3K]1 day ago
7 0
The mass is m=4.03 kg. According to the provided data, we have r = 1.2 m, I = 24 kg·m², N₁ = 40 rpm converting to ω₁= 4.1 rad/s. For N₂ = 32 rpm, ω₂ = 3.3 rad/s. Let's denote the mass of the clay as m kg. The initial angular momentum L₁ is calculated as L₁ = Iω₁, while the final angular momentum L₂ is expressed as L₂= I₂ ω₂ with I₂ indicating the moment of inertia after adding the clay, computed as I₂ = I + m r². Since no external torque is introduced, angular momentum will be conserved, leading to L₁ = L₂. Rearranging gives us Iω₁ = I₂ω₂, or Iω₁ =( I + m r²)ω₂. By inserting the known values of I, r, and ω into the equation, we solve for m, yielding m=4.03 kg.
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Response:

Clarification:

Provided

weight of disk m=12.5 kg

diameter of disc R=0.23 m

weight of ring m_r=7 kg

Force F=9.7 N

N=180 rpm

\omega =\frac{2\pi N}{60}

\omega =6\pi rad/s

Overall moment of inertia

=Disc's moment of inertia +Ring's Moment of Inertia

=0.5\cdot 12.5\times 0.23^2+7\times 0.23^2

=13.25\times 0.23^2=0.7009 kg-m^2

At this point, Torque is T=F\times R=I\cdot \alpha

9.7\times 0.23=0.7\times \alpha

\alpha =3.18 rad/s^2

Utilizing \omega _f=\omega +\alpha t

\omega _f=0 in this scenario

0=6\pi -3.18\times t

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In an inertial frame of reference, a series of experiments is conducted. in each experiment, two or three forces are applied to
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Objects will stay in a stationary position if the total force acting on them amounts to zero; this occurs when equal forces are applied in opposite directions. According to Newton's second law, if the net force on an object is zero, it will not move.
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Answer:

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Explanation:

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by substituting M =0.420kg, g = 9.8m/s^2 into the equation and resolving for F_t we find:

F_t = Mg\:tan(5^o)

F_t = (0.420kg)(9.8m/s^2)\:tan(5^o)

\boxed{F_t = 0.3601N.}

This thrust is linked to the speed of air ejection v through the equation

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where dM/dt signifies the rate of air ejection, which is known to be

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and since 1m^3 = 1.2kg,

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by inserting these values into equation (2), we obtain the value of F_t as:

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Answer:

Explanation:

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P.E = Work

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Dividing both sides by 10 gives:

h = 3600 / 10

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