A. Rank these metals on the basis of their cutoff frequency. Rank from largest to smallest. To rank items as equivalent, overlap them.
B. A yo-yo-shaped device mounted on a horizontal frictionless axis is used to lift a 27 kg box as shown in the figure. The outer radius R of the device is 0.34 m, and the radius r of the hub is 0.11 m. When a constant horizontal force F applied of magnitude 130 N is applied to a rope wrapped around the outside of the device, the box, which is suspended from a rope wrapped around the hub, has an upward acceleration of magnitude 0.75 m/s2. What is the rotational inertia of the device about its axis of rotation
T he Correct Answer and Explanation is :
A. Ranking Metals by Cutoff Frequency
The cutoff frequency (also called threshold frequency) of a metal in the photoelectric effect is directly related to the work function ϕ\phi of the metal. The relationship is: fc=ϕhf_c = \frac{\phi}{h}
Where:
- fcf_c is the cutoff frequency
- ϕ\phi is the work function
- hh is Planck’s constant
Higher work function = higher cutoff frequency
To rank the metals, you’d need their work functions. Assuming the metals are typical photoelectric materials (like Cesium, Sodium, Aluminum, Zinc, and Copper), here are approximate work functions:
| Metal | Work Function (eV) | Cutoff Frequency (Hz) |
|---|---|---|
| Copper | 4.7 | Highest |
| Zinc | 4.3 | |
| Aluminum | 4.1 | |
| Sodium | 2.3 | |
| Cesium | 1.9 | Lowest |
Ranking from Largest to Smallest:
Copper > Zinc > Aluminum > Sodium > Cesium
B. Rotational Inertia of the Yo-Yo Device
Given:
- Mass of the box: m=27 kgm = 27 \, \text{kg}
- Acceleration of the box: a=0.75 m/s2a = 0.75 \, \text{m/s}^2
- Outer radius: R=0.34 mR = 0.34 \, \text{m}
- Hub radius: r=0.11 mr = 0.11 \, \text{m}
- Applied force: F=130 NF = 130 \, \text{N}
- Gravitational acceleration: g=9.8 m/s2g = 9.8 \, \text{m/s}^2
Step-by-Step Solution:
1. Tension in the rope holding the box:
Using Newton’s second law for the box: T=mg−ma=27(9.8−0.75)=27(9.05)=244.35 NT = mg – ma = 27(9.8 – 0.75) = 27(9.05) = 244.35 \, \text{N}
2. Torque Equation:
The device has two torques acting on it:
- From the applied force FF: τF=R⋅F\tau_F = R \cdot F
- From the tension TT: τT=−r⋅T\tau_T = -r \cdot T (negative because it opposes the rotation)
Net torque: τnet=I⋅α=RF−rT\tau_{\text{net}} = I \cdot \alpha = RF – rT
3. Relate angular acceleration α\alpha to linear acceleration:
The rope attached to the hub causes the box to accelerate, so: a=r⋅α⇒α=ara = r \cdot \alpha \Rightarrow \alpha = \frac{a}{r}
4. Plug into the torque equation:
I⋅ar=RF−rT⇒I=r(RF−rT)aI \cdot \frac{a}{r} = RF – rT \Rightarrow I = \frac{r(RF – rT)}{a}
Substitute known values: I=0.11(0.34⋅130−0.11⋅244.35)0.75I = \frac{0.11(0.34 \cdot 130 – 0.11 \cdot 244.35)}{0.75} I=0.11(44.2−26.88)0.75=0.11(17.32)0.75I = \frac{0.11(44.2 – 26.88)}{0.75} = \frac{0.11(17.32)}{0.75} I=1.90520.75≈2.54 kg⋅m2I = \frac{1.9052}{0.75} \approx \boxed{2.54 \, \text{kg} \cdot \text{m}^2}
Explanation
To find the rotational inertia (moment of inertia) of the yo-yo-shaped device, we apply Newton’s laws of motion and torque. The setup involves a rotational system where an external force causes rotation, and that rotation lifts a mass. The outer radius is connected to the applied force while the inner hub radius is connected to the suspended box. Since the box accelerates upward, there is a net force acting on it, indicating that not all of the applied force goes into lifting the box — some of it is stored as rotational energy in the device.
The core idea is to set up a torque balance around the central axis. Two opposing torques act on the device: one from the horizontal force applied at the large radius, and one from the tension in the rope pulling downward at the hub radius. The difference in these torques determines the angular acceleration, which is related to the box’s linear acceleration by α=a/r\alpha = a/r.
By expressing torque as τ=Iα\tau = I\alpha and substituting the relationship between linear and angular acceleration, we rearranged the equation to solve for the rotational inertia II. Using the known values for radii, applied force, tension (derived from Newton’s second law for the box), and acceleration, we calculated the net torque and determined the device’s rotational inertia to be approximately 2.54 kg·m².
This method demonstrates a practical application of rotational dynamics principles, bridging linear motion (the lifting box) with rotational motion (the yo-yo device).
