Answer:
(A) 2.28 × 10⁻³ kg·m²
(B) 3.11 m/s²
(C) 4.38 m/s²
Explanation:
The question presented involves calculating various physical properties related to two metal disks welded together and mounted on a frictionless axis. Part A requires finding the total moment of inertia of the disks. Part B seeks the speed of a weight when released from a certain height, using the smaller disk as a pulley. Part C repeats the setup of Part B but with the string wrapped around the larger disk instead.
Given:
- R₁ = 2.60 cm
- R₂ = 4.90 cm
- M₁ = 0.700 kg
- M₂ = 1.70 kg
- m = 1.50 kg
- d = 1.60 m
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Part (A): Moment of Inertia, I[tex]\hrulefill[/tex]
To answer part A, we will use the formula for the moment of inertia 'I' of a solid disk about an axis through its center, which is given by:
[tex]\boxed{\begin{array}{ccc} \text{\underline{Moment of Inertia of a Solid Disk:}} \\\\ I = \frac{1}{2}mr^2 \\\\ \text{Where:} \\ \bullet \ I \ \text{is the moment of inertia} \\ \bullet \ m \ \text{is the mass of the disk} \\ \bullet \ r \ \text{is the radius of the disk} \end{array} }[/tex]
Thus, we have the equation:
[tex]I_\text{total}=I_1+I_2\\\\\\\\\Longrightarrow I_\text{total} = \dfrac{1}{2}M_1R_1^2+\dfrac{1}{2}M_2R_2^2[/tex]
[tex]\Longrightarrow I_\text{total} = \dfrac{1}{2}(0.700 \text{ kg})(0.0260 \text{ m})^2+\dfrac{1}{2}(1.70 \text{ kg})(0.0490\text{ m})^2[/tex]
[tex]\therefore I_\text{total} \approx \boxed{2.28 \times 10^{-3} \text{ kg$\cdot$m$^2$}}[/tex]
Thus, the total moment of inertia is found of the two disks.
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Part (B): Speed of the Block, with String on Smaller Disk[tex]\hrulefill[/tex]
To answer part B, we will apply Newton's Second Law for both transitional and rotational motion:
[tex]\sum \vec \tau=I \alpha \text{ and } \sum \vec F=ma[/tex]
[tex]\Longrightarrow \underbrace{TR_1}_{\because \ \tau = Fd\sin\theta}=I_{\text{total}} \underbrace{\left(\dfrac{a}{R_1}\right)}_{\because \ a = R\alpha}[/tex]
[tex]\therefore T = \dfrac{I_\text{tot}a}{R_1^2} \dots (1)[/tex]
[tex]\Longrightarrow \vec w - \vec T = ma \dots (2)[/tex]
Plugging (1) into (2):
[tex]\Longrightarrow mg - \dfrac{I_\text{tot}a}{R_1^2} = ma[/tex]
[tex]\Longrightarrow mg = a\left[m + \dfrac{I_\text{tot}}{R_1^2}\right][/tex]
[tex]\therefore a= \dfrac{mg}{\left[m + \dfrac{I_\text{tot}}{R_1^2}\right]}[/tex]
Plug in our given values to determine the acceleration:
[tex]\Longrightarrow a= \dfrac{(1.50 \text{ kg})(9.8 \text{ m/s}^2)}{\left[1.50 \text{ kg} + \dfrac{2.28 \times 10^{-3} \text{ kg$\cdot$m$^2$}}{(0.0260 \text{ m})^2}\right]}\\\\\\\\\therefore a \approx 3.02 \text{ m/s}^2[/tex]
Now using the following kinematic equation we can find the speed of the block just before it hits the ground:
[tex]\Longrightarrow v_f^2=v_0^2+2ad\\\\\\\\\Longrightarrow v_f=\sqrt{(0 \text{ m/s})^2+2(3.02 \text{ m/s}^2)(1.60 \text{ m})}\\\\\\\\\therefore v_f \approx \boxed{ 3.11 \text{ m/s}}[/tex]
Thus, the speed has been found.
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Part (C): Speed of the Block, with String on Larger Disk[tex]\hrulefill[/tex]
To determine the answer to part C, we can follow the same steps but using the larger disk.
[tex]\therefore a= \dfrac{mg}{\left[m + \dfrac{I_\text{tot}}{R_2^2}\right]}[/tex]
[tex]\Longrightarrow a= \dfrac{(1.50 \text{ kg})(9.8 \text{ m/s}^2)}{\left[1.50 \text{ kg} + \dfrac{2.28 \times 10^{-3} \text{ kg$\cdot$m$^2$}}{(0.0490 \text{ m})^2}\right]}\\\\\\\\\therefore a \approx 6.00 \text{ m/s}^2[/tex]
[tex]\Longrightarrow v_f^2=v_0^2+2ad\\\\\\\\\Longrightarrow v_f=\sqrt{(0 \text{ m/s})^2+2(6.00 \text{ m/s}^2)(1.60 \text{ m})}\\\\\\\\\therefore v_f \approx \boxed{ 4.38 \text{ m/s}}[/tex]
Thus, the speed has been found.
