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Andrews [41]
3 years ago
8

Two ladybugs sit on a rotating disk that is slowing down at a constant rate. The ladybugs are at rest with respect to the surfac

e of the disk and do not slip. Ladybug 1 is halfway between ladybug 2 and the axis of rotation. How does the rotational speed of ladybug 1 compare to ladybug 2?
Physics
1 answer:
d1i1m1o1n [39]3 years ago
3 0

The two ladybugs have same rotational (angular) speed

Explanation:

The rotational (angular) speed of an object in circular motion is defined as:

\omega=\frac{\theta}{t}

where

\theta is the angular displacement

t is the time interval considered

Here we have two ladybugs, which are located at two different distances from the axis. In particular, ladybug 1 is halfway between ladybug 2 and the axis of rotation. However, since they rotate together with the disk, and the disk is a rigid body, every point of the disk cover the same angle \theta in the same time t: this means that every point along the disk has the same angular speed, and therefore the two ladybugs also have the same angular speed.

On the other hand, the linear speed of the two ladybugs is different, because it follows the equation:

v=\omega r

where r is the distance from the axis: and since the two ladybugs are located at different r, they have different linear speed.

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

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Answer:1.7 rev/s

Explanation:

Given

Frequency of wheel N_1=2\ rev/s

angular speed \omega_1=2\pi N_1=4\pi\  rad/s

mass of wheel m_1=4.5\ kg

diameter of wheel d_1=0.30\ m=30\ cm

radius of wheel r_1=\frac{d_1}{2}=\frac{30}{2}=15\ cm

mass of clay m_2=2.8\ kg

the radius of the chunk of clay r_2=8\ cm

Moment of inertia of Wheel

I_1=\dfrac{m_1r_1^2}{2}=\dfrac{4.5\times 15^2}{2}\ kg-cm^2

Combined moment of inertia of wheel and clay chunk

I_2=\dfrac{m_1r_1^2}{2}+\dfrac{m_2r_2^2}{2}=\dfrac{4.5\times 15^2}{2}+\dfrac{2.8\times 8^2}{2}\ kg-cm^2

Conserving angular momentum

\Rightarrow I_1\omega_1=I_2\omega_2\\\Rightarrow \dfrac{4.5\times 15^2}{2}\cdot 4\pi=(\dfrac{4.5\times 15^2}{2}+\dfrac{2.8\times 8^2}{2})\omega_2\\\\\Rightarrow \omega _2=\dfrac{4\pi }{1+\dfrac{2.8}{4.5}\times (\dfrac{8}{15})^2}=\dfrac{4\pi}{1+0.1769}=0.849\times 4\pi

Common frequency of wheel and chunk of clay is

\Rightarrow N_2=\dfrac{4\pi \times 0.849}{2\pi}=1.698\approx 1.7\ rev/s

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3 years ago
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Answer:

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A sack of potatoes weighing 16.0-kg falls from a very tall building. At a certain point at the motion downwards, its measured ac
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Answer:

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Explanation:

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