VIPSolutionsPlease solve correct option- Question § © Max score: 100 1 The Teyior tool He equstion used 1o aicuiate the 4 of th tool wih respect to the cuting speed of the tool. Suppose 8 carbide 100s sed wih the cutig speed of / remind he 200i Me of Tin. ¥ the cuting speed reduced by 20% Calcuiie the effect onthe ool We. Assuming machining constert es 350 8nd foo We exponent 2 3 ? Se 5 7 - he to bce by 5 =U © 0 §
Question:
Please solve correct option
- Question § © Max score: 100
1
The Teyior tool He equstion used 1o aicuiate the 4 of th tool wih respect to the cuting speed of the tool. Suppose 8 carbide 100s sed wih the cutig speed of
/ remind he 200i Me of Tin. ¥ the cuting speed reduced by 20% Calcuiie the effect onthe ool We. Assuming machining constert es 350 8nd foo We exponent
2
3
? Se
5
7
- he to bce by
5
=U ©
0 §
This is AI generated Response.
Asked by: Sanjay hipparakar
Created at: 2025-09-18 14:08:42
Asked by: Sanjay hipparakar
Created at: 2025-09-18 14:08:42
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---
# **Question 9**
## **Taylor Tool Life Equation**
The equation is:
\[ VT^n = C \]
where:
- \( V \) = cutting speed
- \( T \) = tool life
- \( n \) = exponent (given as 0.3)
- \( C \) = constant (350, but not needed for ratio problems)
### **Given:**
- Cutting speed reduced by 20% (\( V_{\text{new}} = 0.8V \))
- Find the effect on tool life (\( T_{\text{new}} \) compared to \( T \))
- \( n = 0.3 \)
### **Calculation**
\[
VT^n = (0.8V)T_{\text{new}}^n
\]
\[
V T^n = 0.8V T_{\text{new}}^n
\]
\[
T^n = 0.8 T_{\text{new}}^n
\]
\[
\frac{T^n}{T_{\text{new}}^n} = 0.8
\]
\[
\left(\frac{T}{T_{\text{new}}}\right)^n = 0.8
\]
\[
\frac{T}{T_{\text{new}}} = 0.8^{1/n}
\]
\[
\frac{T}{T_{\text{new}}} = 0.8^{1/0.3} \approx 0.8^{3.333}
\]
Calculate \( 0.8^{3.333} \):
\[
\ln(0.8^{3.333}) = 3.333 \times \ln(0.8) \approx 3.333 \times (-0.2231) = -0.7436
\]
\[
e^{-0.7436} \approx 0.4757
\]
\[
\frac{T}{T_{\text{new}}} \approx 0.4757
\]
\[
T_{\text{new}} \approx \frac{T}{0.4757} \approx 2.1T
\]
So, tool life **increases by about 110%**.
### **Correct Option:**
- **The tool life increased by 110%**
---
# **Question 8**
## **Given:**
- Ball A: \(m_A = 5\,kg,\, u_A = 10\,m/s\)
- Ball B: \(m_B = 10\,kg,\, u_B = -4\,m/s\) (opposite direction)
- Coefficient of restitution \(e = 0.6\)
Let \(v_A\) and \(v_B\) be velocities after impact.
### **Equations:**
#### 1. Conservation of Momentum
\[
m_A u_A + m_B u_B = m_A v_A + m_B v_B
\]
\[
5 \times 10 + 10 \times (-4) = 5 v_A + 10 v_B
\]
\[
50 - 40 = 5 v_A + 10 v_B
\]
\[
10 = 5 v_A + 10 v_B
\]
\[
v_A + 2 v_B = 2
\]
#### 2. Coefficient of Restitution
\[
e = \frac{v_B - v_A}{u_A - u_B}
\]
\[
0.6 = \frac{v_B - v_A}{10 - (-4)} = \frac{v_B - v_A}{14}
\]
\[
v_B - v_A = 0.6 \times 14 = 8.4
\]
\[
v_B = v_A + 8.4
\]
### **Solving the two equations:**
Substitute \(v_B\) from above into the first equation:
\[
v_A + 2(v_A + 8.4) = 2
\]
\[
v_A + 2v_A + 16.8 = 2
\]
\[
3v_A = 2 - 16.8 = -14.8
\]
\[
v_A = -4.93\, m/s
\]
\[
v_B = v_A + 8.4 = -4.93 + 8.4 = 3.47\, m/s
\]
### **Correct Option:**
- **3.47 m/s and -4.93 m/s**
---
## **Final Answers:**
### **Question 9:**
**The tool life increased by 110%.**
### **Question 8:**
**3.47 m/s and -4.93 m/s.**
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