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nataly862011 [7]
3 years ago
8

How do inertia and centripetal force combine to keep an object moving in circular motion?

Physics
2 answers:
katen-ka-za [31]3 years ago
6 0
As the centripetal force<span> acts upon an </span>object moving <span>in a </span>circle<span> at constant speed, the </span>force<span> always acts inward as the velocity of the </span>object<span> is directed tangent to the </span>circle. ... In fact, whenever the unbalanced centripetal force<span> acts perpendicular to the direction of </span>motion<span>, the speed of the </span>object will<span> remain constant.</span>
mash [69]3 years ago
5 0

Answer:

As we know that inertia it the property of mass which resist any change in the state of motion of the object. Which means object has tendency to move in its initial state of motion and when we try to change that state of motion then object will resist to change that.

While moving in circular path the object will have tendency to change it's direction of motion at each instant. So this change in the direction of motion at every instant is due to centripetal acceleration of the object.

Now when object is revolving in circular path then the inertia of the object will help to move the object in tangential direction at every moment of the motion while the centripetal acceleration of the object will have tendency to change the direction at each moment of the motion.

So here we will say that inertia and centripetal force combine to keep an object moving in circular motion

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You have a two-wheel trailer that you pull behind your ATV. Two children with a combined mass of 77.2 kg hop on board for a ride
lions [1.4K]

To solve this problem it is necessary to apply the kinematic equations of motion and Hook's law.

By Hook's law we know that force is defined as,

F= kx

Where,

k = spring constant

x = Displacement change

PART A) For the case of the spring constant we can use the above equation and clear k so that

k= \frac{F}{x}

k = \frac{mg}{x}

k= \frac{77.2*9.8}{0.0637}

k = 11876.92N/m

Therefore the spring constant for each one is 11876.92/2 = 5933.46N/m

PART B) In the case of speed we can obtain it through the period, which is given by

T = \frac{2\pi}{\omega}

Re-arrange to find \omega,

\omega = \frac{2\pi}{T}

\omega = \frac{2\pi}{2.14}

\omega = 2.93rad/s

Then through angular kinematic equations where angular velocity is given as a function of mass and spring constant we have to

\omega^2 = \frac{k}{m}

m = \frac{k}{\omega^2}

m = \frac{ 11876.92}{2.93}

m = 4093.55Kg

Therefore the mass of the trailer is 4093.55Kg

PART C) The frequency by definition is inversely to the period therefore

f = \frac{1}{T}

f = \frac{1}{2.14}

f = 0.4672 Hz

Therefore the frequency of the oscillation is 0.4672 Hz

PART D) The time it takes to make the route 10 times would be 10 times the period, that is

t_T = 10*T

t_T = 10 *2.14s

t_T = 21.4s

Therefore the total time it takes for the trailer to bounce up and down 10 times is 21.4s

5 0
3 years ago
Which wave must have a medium to travel?
o-na [289]

Letter B

without a medium, there is nothing to compress, hence, no wave. A fast- medium like a gas (air) is easy to compress and allows waves to move through it easily. a slow medium, like a liquid, is still pretty fast, but not as fast as air.

6 0
3 years ago
Read 2 more answers
Peg P is driven by the forked link OA along the path described by r = eu, where r is in meters. When u = p4 rad, the link has an
8_murik_8 [283]

Answer:

The transverse component of acceleration is 26.32 m/s^2 where as radial the component of acceleration is 8.77 m/s^2

Explanation:

As per the given data

u=π/4 rad

ω=u'=2 rad/s

α=u''=4 rad/s

r=e^u

So the transverse component of acceleration are given as

a_{\theta}=(ru''+2r'u')\\

Here

r=e^u\\r=e^{\pi/4}\\r=2.1932 m

r'=e^u.u'\\r'=2.1932 \times 2\\r'=4.3864 m

So

a_{\theta}=(ru''+2r'u')\\a_{\theta}=(2.1932\times 4+2\times 4.3864 \times 2)\\a_{\theta}=26.32 m/s\\

The transverse component of acceleration is 26.32 m/s^2

The radial component is given as

a_r=r''-r\theta'^2

Here

r''=e^u.u'^2+e^u u''\\r''=2.1932 \times (2)^2+2.1932\times 4\\r''=17.5456 m

So

a_r=r''-ru'^2\\a_r=17.5456-2.1932\times (2)^2\\a_r=8.7728 m/s^2

The radial component of acceleration is 8.77 m/s^2

6 0
3 years ago
QUESTION 9
Alik [6]

Answer:

The answer is option a.

Hope this helps

4 0
3 years ago
A bug is sitting on the edge of a rotating disk. At what angular velocity will the bug slide off the disk if its radius is 0.241
Ivahew [28]

Answer:

ω = 3.61 rad/sec

Explanation:

Firstly, we should know that the bug will not slip if friction can provide sufficient opposing force.

μmg = mv^2/r = mω^2r

Thus;

μg = ω^2r

ω^2 = μg/r

ω = √(μg/r)

ω = √(0.321 * 9.8)/0.241

ω = √(13.05)

= 3.61 rad/sec

3 0
3 years ago
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