Please help on this one?

A solid sphere, solid cylinder, and a hollow pipe all have equal masses and radii. If the three of them are released simultaneou

Roman55 [17]

Answer:

The solid sphere will reach the bottom first.

Explanation:

In order to develop this problem and give it a correct solution, it is necessary to collect the concepts related to energy conservation. To apply this concept, we first highlight the importance of conserving energy so we will match the final and initial energies. Once this value has been obtained, we will concentrate on finding the speed, and solving what is related to the Inertia.

In this way we know that,

$\Delta KE = - \Delta PE$

$KE_t + KE_r = mgh$

We know as well that the lineal and angular energy are given by,

$KE_r = \frac{1}{2}I\omega^2$

And the tangential kinetic energy as

$KE_t = \frac{1}{2} mv^2$

Where $\omega = \frac{v}{R}$

Replacing

$\frac{1}{2}mv^2 + \frac{1}{2}I\frac{v}{R} = mgh$

Re-arrange for v,

$v=\sqrt{\frac{2mgh}{m+I/R^2}}$

We have here three different objects: solid cylinder, hollow pipe and solid sphere. We need the moment inertia of this objects and replace in the previous equation found, then,

For hollow pipe:

$I_{hp}=mR^2$

$v_{hp}=\sqrt{\frac{2mgh}{m+(mR^2)/R^2}}$

$v_{hp}=\sqrt{\frac{2mgh}{m+m)}$

$v_{hp}=\sqrt{gh}$

For solid cylinder:

$I_{sc}=\frac{1}{2}mR^2$

$v_{sc}=\sqrt{\frac{2mgh}{m+(1/2mR^2)/R^2}}$

$v_{sc}=\sqrt{\frac{2mgh}{m+1/2m}}$

$v_{sc}=\sqrt{\frac{3}{4}gh}$

For solid sphere,

$I_{ss}=\frac{2}{5}mR^2$

$v_{ss}=\sqrt{\frac{2mgh}{m+(2/5mR^2)/R^2}}$

$v_{ss}=\sqrt{\frac{2mgh}{m+2/5m}}$

$v_{ss}=\sqrt{\frac{10}{7}gh}$

Then comparing the speed of the three objects we have:

$v_{hp}$

$\sqrt{gh}$

3 0

4 years ago

Tropism to stimuli could be slow at fast but all plants respond to __?

FromTheMoon [43]

HI MY NAME IS GUNTA!!!

5 0

4 years ago

A dumbbell-shaped object is composed by two equal masses, m, connected by a rod of negligible mass and length r. If I1 is the mo

Sati [7]

Answer:

The correct answer to the following question will be Option A (I1 > I2).

Explanation:

Method for moment of inertia because of it's viewpoint including object at a mean distance "r" from the axis is,

⇒ mr²

For Case 1:

Let the length of a rod be "r".

The axis passes via the middle of that same rod so that the range from either the axis within each dumbbell becomes " $\frac{r}{2}$ ".

Now,

Now total moment of inertia = sum of inertial moment due to all of the dumbbell

⇒ $l1=m(\frac{r}{2})^2+m(\frac{r}{2})^2$

⇒ $\frac{mr^2}{2}$

For Case 2:

Axis moves via one dumbbell because its range from either the axis becomes zero (0) and its impact is zero only at inertia as well as other dumbbell seems to be at a range "r" from either the axis

Now,

Total moment of inertia = moment of inertia of dumbbell at distance "r".

$l2=mr^2$

And now we can infer from this one,

⇒ $mr^2>\frac{mr^2}{2}$

So that "I1 > I2" is the right answer.

5 0

4 years ago

Why do some balloons pop when they are left in sunlight for too long

Ghella [55]

the sun raises the temperatures of molecules inside the balloon. As the molecules receives more kinetic energy, they move apart from each other and expand in volume. Depend on the balloon's material, it might not be strong enough to handle the pressure exerted by the gas inside, and the balloon will pop.

4 0

4 years ago

Which of the following items involve a wedge?

elena55 [62]

Answer:

D

Explanation:

Maybe your question is meant to be: which is not a wedge, because wedge is a combination of two inclined planes, use to separate bodies which are held together by large forces. Option A,B & C are all wedge except D.

8 0

3 years ago

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