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Brums [2.3K]
3 years ago
7

Draw velocity-ime graph for uniform moion of an object , when initially body is at rest.

Physics
1 answer:
Klio2033 [76]3 years ago
8 0
When the body is at rest, its speed is zero, and the graph lies on the x-axis.

When the body is in uniform motion, the speed is constant, and the graph is a horizontal line, parallel to the x-axis and some distance above it.

It's impossible to tell, based on the given information, how these two parts of the
graph are connected.  There must be some sloping (accelerated) portion of the graph
that joins the two sections, but it cannot be accounted for in either the statement
that the body is at rest or that it is in uniform motion, since acceleration ... that is,
any change of speed or direction ... is not 'uniform' motion'.
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A 240 N sphere 0.20 m in radius rolls, without slipping 6.0 m downa
Shalnov [3]

Answer:

The angular speed of the sphere at the bottom of the hill is 31.39 rad/s.

Explanation:

It is given that,

Weight of the sphere, W = 240 N

Radius of the sphere, r = 0.2 m

Angle with the horizontal, \theta=28^{\circ}

We need to find the angular speed of the sphere at the bottom of the hill if it starts  from rest.

As per the law of conservation of energy, the total energy at the top is equal to the energy at the bottom.

Gravitational energy = translational energy + rotational energy

So,

mgh=\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2

I is the moment of inertia of the sphere, I=\dfrac{2}{5}mr^2

Also, v=r\omega

h is the height of the ramp, h=l\ sin\theta

mgl\ sin\theta=\dfrac{1}{2}m(r\omega)^2+\dfrac{1}{2}I\omega^2

On solving the above equation we get :

\omega=\sqrt{\dfrac{10gl\ sin\theta}{7r^2}}

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So, the angular speed of the sphere at the bottom of the hill is 31.39 rad/s. Hence, this is the required solution.

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