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If a sphere slides down from rest on an inclined plane of hypotenuse $L$ and height $H$ : Work done by friction is converted into rolling kinetic energy and as well as heat energy.

If $f$ is the friction force then is work done by friction $fL = $Heat energy $+ $ rolling $KE$ ? Or does the formula $fL $ gives only the heat energy lost by friction force ?

My second question regarding this is : Is it correct to write work- energy theorem at the feet of the inclined plane as :

$$KE\ _{linear}=mgh+fl$$

Danny LeBeau
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2 Answers2

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If you are considering kinetic energy only due to linear motion . The your equation is wrong . It's : $$ {KE\ _{linear}} + fl = mgh $$

Because friction is working against the linear motion.

Bhavay
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Friction and rolling This not as banal a problem as you may expect at first sight.

First, study the emerging rotational motion:

$$F_N=mg\cos\theta$$ $$F_f=\mu_k F_N=\mu_k mg\cos\theta$$ Torque about the axis of rotation causes angular acceleration: $$\tau=I\alpha$$ $$F_f R=I\frac{\text{d}\omega}{\text{d}t}$$ $$\mu_k mg\cos\theta R=\gamma mR^2 \frac{\text{d}\omega}{\text{d}t}$$ where $\gamma$ is a coefficient depending on the exact shape of the rotating body. $$\frac{\text{d}\omega}{\text{d}t}=\frac{\mu_kg\cos\theta}{\gamma R}$$ Assuming $\omega =0$ at $t=0$: $$\omega(t)=\frac{\mu_kg\cos\theta}{\gamma R}t$$ Now study the translational motion: $$F_s-F_f=ma$$ $$mg\sin\theta-\mu_k mg\cos\theta=ma$$ $$\frac{\text{d}v}{\text{d}t}=g(\sin\theta-\mu_k \cos\theta)$$ Assuming $v=0$ at $t=0$: $$v(t)=g(\sin\theta-\mu_k \cos\theta)t$$ The object reaches rolling without slipping (pure rolling) when:

$$v(t)=\omega(t)R$$

which with some substituting and reworking gives the relationship:

$$\mu_k=\frac{\gamma}{\gamma+1}\tan\theta$$

So how to calculate the relevant energies?

You already know the work done by the friction force.

How much energy is used to get the object to roll? Calculate the time needed to reach the bottom of the incline ($0 \to L$) and from there calculate $\omega(t)$ and use that to calculate the change in rotational kinetic energy.

I hope this helps.

Gert
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  • Can you please explain if The sphere is only sliding during the motion , that is $v \ne \omega r $ and i use the formula Work done by friction $fL$ , it means the total work done by friction(that's what i think you mean) or just the heat loss by frictional force ? – Danny LeBeau Aug 03 '20 at 18:01
  • It's only sliding when $v \neq \omega R$, correct. After that, during pure rolling, $F_f$ no longer does work. That's why you need to determine the point $t$ where $v=\omega R$. – Gert Aug 03 '20 at 18:11
  • Sir I am only talking about the case when during the whole motion the sphere slides and never enters pure rolling motion. – Danny LeBeau Aug 03 '20 at 18:14
  • Well, that's fine, no problemo! In that case calculate all the energies and know that the total energy is the change in potential energy: $\Delta U=\Delta E_{trans}+\Delta E_{rot}+\Delta Q_{heat}$. – Gert Aug 03 '20 at 18:18
  • And can we write $\Delta E_{rot} + \Delta Q_{heat}=fl$? I just want to be absolutely sure. – Danny LeBeau Aug 03 '20 at 18:22
  • Potential energy ($mgh$) is converted into rotational kinetic energy, translational kinetic energy and 'other' (mostly heat, some noise, etc). So it's $\Delta U=\Delta E_{trans}+\Delta E_{rot}+\Delta Q_{heat}$. The three first terms can be calculated exactly, so the fourth can be deduced from that. – Gert Aug 03 '20 at 18:27
  • Okay...but if i use the formula $fl$ and get the answer as $X \ J$,Will $X \ J$ joules be equal to $\Delta Q_{heat} + \Delta E_{rot} $? Just answer it in yes or no to avoid any further complications . – Danny LeBeau Aug 03 '20 at 18:33
  • Sorry, I misunderstood you there for a second. The answer to your last question is Yes. – Gert Aug 03 '20 at 18:35
  • Thanks for the upvote. – Gert Aug 03 '20 at 18:44