Question

gives the speed v versus time t for a 0.500 kg object of radius 6.00 cm that rolls smoothly down a 30° ramp. The scale on the velocity axis is set by vs = 4.0 m/s. What is the rotational inertia of the object?

Answer 1

Answer:

Rotational inertia of the object is given as

$I = 7.2 \times 10^{-4} kg m^2$

Explanation:

As we know that the acceleration of the object on inclined plane is given as

$a = \frac{gsin\theta}{1 + k^2/R^2}$

now we know that velocity at any instant of time is given as

$v = at$

now we know that if the graph between velocity and time is given then the slope of the graph will be same as acceleration

so here we have

$\frac{gsin\theta}{1 + k^2/R^2} = slope$

now from the graph slope of the graph is given as

$slope = \frac{3.5 - 0}{1}$

$\frac{gsin\theta}{1 + k^2/R^2} = 3.5$

$\frac{9.81 sin30}{1 + k^2/R^2} = 3.5$

$k^2 = 0.4 R^2$

now rotational inertia is given as

$I = mk^2$

$I = 0.5(0.4)(0.06)^2$

$I = 7.2 \times 10^{-4} kg m^2$

Answer 2

The rotational inertia of the object when the speed v versus time t for a 0.500 kg should be I = 7.2*10^-4 kgm^2.

Calculation of the rotational inertia:

Since speed v versus time t for a 0.500 kg object of radius 6.00 cm that rolls smoothly down a 30° ramp.

Slope = 3.5-0/1

gsin theta / 1 + k^2/R^2 = 3.5

0.81 sin30 / 1 + k^2 / R^2= 3.5

K^2 = 0.4R^2

Now finally

I = mk^2

= 0.5(0.4)(0.06)^2

= 7.2*10^-4 kgm^2.

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