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# Tutor profile: Erin H.

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Erin H.
PhD Candidate, Physics
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## Questions

### Subject:Physics (Thermodynamics)

TutorMe
Question:

$$N=H$$ molecules of an ideal diatomic gas are confined to a volume $$V$$ that is maintained at constant temperature $$T$$. At the temperature $$T$$, each molecule has five degrees of freedom (3 translational and 2 rotational). (1) What is the internal energy of the gas? (2) Find the heat capacities: $$C_V$$, $$C_P$$, and $$\gamma$$ The gas is irradiated with ultraviolet light and one third of the $$N$$ molecules dissociate, each into two atoms. (3) What is the internal energy of the gas? (4) Find the heat capacities: $$C_V$$, $$C_P$$, and $$\gamma$$

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Erin H.

1. The molecules have five degrees of freedom. That means our energy is $$E_0 = \frac{\alpha}{2}NkT = \frac{5}{2}HkT$$ 2. $$C_V = \Big(\frac{dE}{dT}\Big)_V = \frac{5}{2}Hk$$ $$C_P = C_V + Nk = \frac{7}{2}Hk$$ $$\gamma = \frac{C_P}{C_V} = \frac{7}{5}$$ 3. Two thirds of the molecules still have 5 degrees of freedom: $$E_2 = \frac{5}{2}\Big(\frac{2}{3}H\Big) kT = \frac{5}{3} HkT$$ One third of the diatomic molecules split - we now have $\frac{2}{3} H$ monoatomic particles which each have three degrees of freedom. $$E_1 = \frac{3}{2}\Big(\frac{2}{3}H\Big) kT = HkT$$ Our total energy is the sum of the two: $$E = E_1 + E_2 = \frac{8}{3}HkT$$ 4. We calculate our heat capacities the same way as before $$C_V = \Big(\frac{dE}{dT}\Big)_V = \frac{8}{3}Hk$$ $$C_P = C_V + Nk = C_V + \frac{4}{3}Hk = \frac{12}{3}Hk$$ $$\gamma = \frac{C_P}{C_V} = \frac{3}{2}$$

### Subject:Physics (Newtonian Mechanics)

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Question:

A merry-go-round with a moment of inertia equal to 1260 kg$$\cdot$$m$$^2$$ and a radius of 2.5 m rotates with negligible friction at 1.70 rad/s. A child initially standing still next to the merry-go-round jumps onto the edge of the platform straight toward the axis of rotation causing the platform to slow to 1.25 rad/s. What is her mass?

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Erin H.

FIND Mass of the girl KNOW: I = 1260 $$kg\cdot m^2$$ r = 2.5 m $$\omega_1$$ = 1.70 rad/sec $$\omega_2$$ = 1.25 rad/sec L = I$$\omega$$ PLAN & SOLVE Without too much work we can say that the change in angular momentum of the merry-go-round has to be equal to the change in angular momentum of the girl. Therefore $$\Delta L = \Delta L_g$$ $$I(\omega_1 - \omega_2) = m r^2 \omega_2$$ $$m = \frac{I(\omega_1 - \omega_2)}{r^2\omega_2}$$ $$m = 72.6 kg$$ Which is about 140 lbs. This is big for a child but not unreasonable.

### Subject:Physics

TutorMe
Question:

A 230-kg beam 2.7 m in length slides broadside down the ice with a speed of 18 m/s. A 65 kg man at rest grabs one end as it goes past and hangs on as both he and the beam go spinning down the ice. Assume frictionless motion. (a) How fast does the center of mass of the system move after the collision? (b) With what angular velocity does the system rotate about its center of mass?

Inactive
Erin H.

FIND: (a) velocity of the center of mass (b) angular velocity of the center of mass KNOW: M = 230 kg L = 2.7 m $$v_0$$ = 18 m/s m = 65 kg PLAN & SOLVE (a) This part is simple, which is why it's only 5 points. We treat the beam and the man as separate objects at their own center of mass and calculate as if it's an inelastic collision. \begin{eqnarray*} m_1 v_1 + m_2 v_2 &=& (m_1 + m_2)v' \\ Mv_0 &=& (M + m)v'\\ v' &=& \frac{M v_0}{M + m}\\ &=& 14.03\ m/s \end{eqnarray*} This makes sense - it's less than the original velocity and is about 28 mph which isn't unreasonable. (b) This one is a little more complicated, so it's 10 points. First of all we have to realize that the center of mass changes with the addition of the man's mass. Setting my zero at the old center of mass of the beam, I can calculate the center of mass just after the collision: $$d = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} = \frac{m L/2}{M + m} = 0.3 m$$ So my axis of rotation is now going to be 0.3 m from the old center of rotation towards the man. We realize now that our problem is no longer symmetric, so we need to use the parallel axis theorem to find the moment of inertia: $$I = \frac{1}{12}ML^2 + Md^2 + m(L/2 - d)^2$$ We then compare the angular momentum ***about the new axis*** before and after the collision: \begin{eqnarray*} L_1 &=& L_2 \\ Mv_0 d &=& (\frac{1}{12}ML^2 + Md^2 + m(L/2 - d)^2)\omega \\ &&\\ \omega &=& \frac{Mv_0 d}{\frac{1}{12}ML^2 + Md^2 + m(L/2 - d)^2}\\ &=& 5.3\text{ rad/sec} \end{eqnarray*}

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