Question

You live on a planet far from ours. "Based on extensive communication with a physicist on earth", you have determined that all laws of physics on your planet are the same as ours and you have adopted the same units of seconds and meters as on earth. But you suspect that the value of g, the acceleration of an object in free fall near the surface of your planet, is different from what it is on earth. To test this, you take a solid uniform cylinder and let it roll down an incline. The vertical height h of the top of the incline above the lower end of the incline can be varied. You measure the speed vcm of the center of mass of the cylinder when it reaches the bottom for various values of h. You plot v2cm (in m2/s2) versus h (in m) and find that your data lie close to a straight line with a slope of 6.42 m/s2

Answer 1

Answer:

8.56 m/s2

Explanation:

Using law of energy conservation while taking into account of the rotational and translation kinetic energy, when the solid cylinder rolls down the incline we have the potential energy converted to kinetic energy:

$E_p = E_{kv} + E_{k\omega}$

$mgh = mv^2/2 + I\omega^2/2$

where m is the mass, $I = mr^2/2$ is the moments of inertia of the solid cylinder $\omega = v / r$ is the angular speed of the cylinder

$mgh = mv^2/2 + \frac{mr^2}{2}\frac{(v/r)^2}{2}$

$mgh = mv^2/2 + mv^2/4 = 3mv^2/4$

$h = 3v^2/(4g)$

So if you plot a liner chart of h vs $v^2$ and get a slope of 6.42 then that means 3/(4g) = 6.42 so $g = 6.42*4/3 = 8.56 m/s^2$

The gravitational acceleration on this planet is 8.56 m/s2

Answer 2

According to the slope of the graph, the gravitational acceleration on the given planet is 4.815 m/s².

Finding the acceleration due to gravity:

Let the mass of the cylinder be m,

the height of the incline be h, and the final velocity of the cylinder is v.

At the top of the incline, the cylinder has only gravitational potential energy, at the bottom the height is zero so potential energy is zero, by the cylinder has velocity at the bottom.

So, according to the law of conservation of energy, the potential energy of the rod is converted into translational and rotational kinetic energies, given by:

mgh = ¹/₂mv² + ¹/₂Iω²

now, ω = v/r and I = ¹/₂mr² is the moment of inertia of a solid cylinder

so,

mgh = ¹/₂mv² + ¹/₂× ¹/₂mr²(v/r)²

gh = ¹/₂v² + ¹/₄v²

gh = (3/4)v²

g = (3/4)(v²/h)

it is given that, v²/h = 6.42 m/s²

g = (3/4)×6.42 m/s²

g = 4.815 m/s²

Learn more about law of conservation of energy:

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