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Two metal disks, one with radius R1
= 2.60 cm
and mass M1
= 0.700 kg
and the other with radius R2
= 4.90 cm
and mass M2
= 1.70 kg
, are welded together and mounted on a frictionless axis through their common center Two metal disks are welded together and mounted on the horizontal axis through their common center. The smaller disk of radius R 1 is to the left of the larger disk of radius R 2. The disks are in contact. The smaller disk has a block of mass 1.50 kilograms suspended from it by a vertical string wrapped around its edge.

Part B
A light string is wrapped around the edge of the smaller disk, and a 1.50 kg
block is suspended from the free end of the string. If the block is released from rest at a distance of 1.60 m
above the floor, what is its speed just before it strikes the floor?

Part C
Repeat part B
, this time with the string wrapped around the edge of the larger disk.​

Two metal disks, one with radius R1 = 2.60 cm and mass M1 = 0.700 kg and the other with radius R2 = 4.90 cm and mass M2 = 1.70 kg , are welded together and moun class=

Verified Answer :

Answer:

## Analyzing the Falling Block System with Disks (Part B & C)

We can analyze this system using concepts of conservation of mechanical energy and rotational motion. Let's denote the following:

g: acceleration due to gravity (9.81 m/s²)

h: initial height of the block (1.60 m)

v: speed of the block just before hitting the floor (unknown)

I: total moment of inertia of the combined disks (depends on which disk the string is wrapped around)

ω: angular velocity of the disks (related to v)

Part B: String Wrapped Around Smaller Disk (R1)

1. Initial Mechanical Energy: The initial energy consists solely of the gravitational potential energy of the block.

PE_initial = mgh = (1.50 kg)(9.81 m/s²)(1.60 m) = 23.54 J

2. Final Mechanical Energy: Just before hitting the floor, the energy is converted to kinetic energy of the block and rotational kinetic energy of the disks.

PE_final = KE_block + KE_rotational

KE_block = 1/2 mv^2 (where m is the block's mass)

KE_rotational = 1/2 I ω^2 (where I depends on the moment of inertia of the smaller disk and larger disk combined, considering their masses and radii)

3. Conservation of Mechanical Energy: Apply the principle of conservation of mechanical energy.

PE_initial = PE_final

23.54 J = 1/2 mv^2 + 1/2 I ω^2

4. Relating ω and v: Since the string is wrapped around the smaller disk (radius R1), the linear velocity (v) of the block relates to the angular velocity (ω) of the disks:

v = R1 * ω

5. Substituting and Solving: Substitute the relationship between v and ω into the conservation equation and solve for v. This will involve finding the moment of inertia (I) for the smaller and larger disk combination, which depends on specific formulas based on their masses and radii.

Part C: String Wrapped Around Larger Disk (R2)

Repeat the same analysis as Part B, but consider the moment of inertia (I) for the combination where the string is wrapped around the larger disk (radius R2). This will result in a different final equation for v due to the change in I.

Solving for v in both parts requires knowledge of the specific formulas for calculating the moment of inertia for the combined disks depending on the smaller or larger disk being involved. Once you have the equations with I for each case, you can solve for v numerically.

Rochirion

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

\(\hrulefill\)

Part (A): Moment of Inertia, I\(\hrulefill\)

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:

\(\boxed{\begin{array}{ccc} \text{\underline{Moment of Inertia of a Solid Disk:}} \\\\ I = \frac{1}{2}mr^2 \\\\ \text{Where:} \\ \bullet \ I \ \text{is the moment of inertia} \\ \bullet \ m \ \text{is the mass of the disk} \\ \bullet \ r \ \text{is the radius of the disk} \end{array} }\)

Thus, we have the equation:

\(I_\text{total}=I_1+I_2\\\\\\\\\Longrightarrow I_\text{total} = \dfrac{1}{2}M_1R_1^2+\dfrac{1}{2}M_2R_2^2\)

   

\(\Longrightarrow I_\text{total} = \dfrac{1}{2}(0.700 \text{ kg})(0.0260 \text{ m})^2+\dfrac{1}{2}(1.70 \text{ kg})(0.0490\text{ m})^2\)

\(\therefore I_\text{total} \approx \boxed{2.28 \times 10^{-3} \text{ kg$\cdot$m$^2$}}\)

Thus, the total moment of inertia is found of the two disks.

\(\hrulefill\)

Part (B): Speed of the Block, with String on Smaller Disk\(\hrulefill\)

To answer part B, we will apply Newton's Second Law for both transitional and rotational motion:

\(\sum \vec \tau=I \alpha \text{ and } \sum \vec F=ma\)

\(\Longrightarrow \underbrace{TR_1}_{\because \ \tau = Fd\sin\theta}=I_{\text{total}} \underbrace{\left(\dfrac{a}{R_1}\right)}_{\because \ a = R\alpha}\)

\(\therefore T = \dfrac{I_\text{tot}a}{R_1^2} \dots (1)\)

\(\Longrightarrow \vec w - \vec T = ma \dots (2)\)

Plugging (1) into (2):

\(\Longrightarrow mg - \dfrac{I_\text{tot}a}{R_1^2} = ma\)

\(\Longrightarrow mg = a\left[m + \dfrac{I_\text{tot}}{R_1^2}\right]\)

\(\therefore a= \dfrac{mg}{\left[m + \dfrac{I_\text{tot}}{R_1^2}\right]}\)

Plug in our given values to determine the acceleration:

\(\Longrightarrow a= \dfrac{(1.50 \text{ kg})(9.8 \text{ m/s}^2)}{\left[1.50 \text{ kg} + \dfrac{2.28 \times 10^{-3} \text{ kg$\cdot$m$^2$}}{(0.0260 \text{ m})^2}\right]}\\\\\\\\\therefore a \approx 3.02 \text{ m/s}^2\)

Now using the following kinematic equation we can find the speed of the block just before it hits the ground:

\(\Longrightarrow v_f^2=v_0^2+2ad\\\\\\\\\Longrightarrow v_f=\sqrt{(0 \text{ m/s})^2+2(3.02 \text{ m/s}^2)(1.60 \text{ m})}\\\\\\\\\therefore v_f \approx \boxed{ 3.11 \text{ m/s}}\)

Thus, the speed has been found.

\(\hrulefill\)

Part (C): Speed of the Block, with String on Larger Disk\(\hrulefill\)

To determine the answer to part C, we can follow the same steps but using the larger disk.

\(\therefore a= \dfrac{mg}{\left[m + \dfrac{I_\text{tot}}{R_2^2}\right]}\)

\(\Longrightarrow a= \dfrac{(1.50 \text{ kg})(9.8 \text{ m/s}^2)}{\left[1.50 \text{ kg} + \dfrac{2.28 \times 10^{-3} \text{ kg$\cdot$m$^2$}}{(0.0490 \text{ m})^2}\right]}\\\\\\\\\therefore a \approx 6.00 \text{ m/s}^2\)

\(\Longrightarrow v_f^2=v_0^2+2ad\\\\\\\\\Longrightarrow v_f=\sqrt{(0 \text{ m/s})^2+2(6.00 \text{ m/s}^2)(1.60 \text{ m})}\\\\\\\\\therefore v_f \approx \boxed{ 4.38 \text{ m/s}}\)

Thus, the speed has been found.

View image Rochirion

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