A cylindrical bar with 200 mm diameter is being turned with a tool having geometry 0° - 9° - 7° - 8° - 15° - 30° - 0.05 inch (Coordinate system, ASA) resulting in a cutting force F_c1. If the tool geometry is changed to 0° - 9° - 7° - 8° - 15° - 0° - 0.05 inch (Coordinate system, ASA) and all other parameters remain unchanged, the cutting force changes to F_c2. Specific cutting energy (in J/mm³) is U_c = U_o (t₁)-0.4, where Uo is the specific energy coefficient, and t1 is the uncut thickness in mm. The value of percentage change in cutting force F_c2, i.e. $\left(\frac{F_{c2}-F_{c1}}{F_{c1}}\right)\times 100,$ is ______ (round off to one decimal place)

The correct answer is -5.5

Step-by-Step Explanation:

First, let's understand the relation between specific cutting energy ($U_c$) and cutting force ($F_c$). Specific cutting energy is defined as the cutting energy consumed per unit volume of material removed, which can be expressed in terms of the cutting force, feed rate ($f$), and depth of cut ($d$) as:
$$U_c=\frac{F_c}{f\times d}$$

We are given the empirical relation for specific cutting energy:
$$U_c=U_o{\left(t_1\right)}^{-0.4}$$
where $t_1$ is the uncut chip thickness in mm, and $U_o$ is the specific energy coefficient.

In orthogonal or oblique turning, the uncut chip thickness $t_1$ is related to the feed rate $f$ and the principal cutting edge angle (or tool approach angle) $\lambda$ by the formula:
$$t_1=f\times sin\lambda$$

Substituting $t_1$ into the specific cutting energy equation, we obtain:
$$\frac{F_c}{f\times d}=U_o{(f\times sin\lambda )}^{-0.4}$$

Since the feed rate $f$, depth of cut $d$, and specific energy coefficient $U_o$ remain constant during the change of tool geometry, we can set up a proportionality relation for the cutting force:
$$F_c\propto {(sin\lambda )}^{-0.4}$$

Now, let's identify the tool approach angle $\lambda$ for both tool geometries. Under the ASA (American Standards Association) coordinate system, the tool signature is designated as:

Back Rake Angle - Side Rake Angle - End Relief Angle - Side Relief Angle - End Cutting Edge Angle - Side Cutting Edge Angle ($C_s$) - Nose Radius
From the given tool geometries:
1. First tool: 0° - 9° - 7° - 8° - 15° - 30° - 0.05 inch ⇒ Side Cutting Edge Angle $C_{s1}={30}^{ˆ}$
2. Second tool: 0° - 9° - 7° - 8° - 15° - 0° - 0.05 inch ⇒ Side Cutting Edge Angle $C_{s2}=0^{ˆ}$

The relation between the principal cutting edge angle $\lambda$ and the side cutting edge angle $C_s$ is:
$$\lambda ={90}^{ˆ}-C_s$$
Therefore, we calculate ${\lambda }_1$ and ${\lambda }_2$ as:
$${\lambda }_1={90}^{ˆ}-{30}^{ˆ}={60}^{ˆ}$$
$${\lambda }_2={90}^{ˆ}-0^{ˆ}={90}^{ˆ}$$

The percentage change in the cutting force is given by:
$$\text{Percentage Change}=\left(\frac{F_{c2}-F_{c1}}{F_{c1}}\right)\times 100=[{\left(\frac{sin{\lambda }_2}{sin{\lambda }_1}\right)}^{-0.4}-1]\times 100$$

Substitute the value of the angles:
$$\text{Percentage Change}=[{\left(\frac{sin{90}^{ˆ}}{sin{60}^{ˆ}}\right)}^{-0.4}-1]\times 100$$
Since $sin{90}^{ˆ}=1$ and $sin{60}^{ˆ}=\frac{\sqrt{3}}{2}\approx 0.8660$:
$$\frac{sin{90}^{ˆ}}{sin{60}^{ˆ}}=\frac{1}{0.8660}\approx 1.1547$$
Now calculate the exponent:
$${\left(1.1547\right)}^{-0.4}\approx 0.9441$$

Subtract 1 and multiply by 100 to get the percentage value:
$$\text{Percentage Change}=(0.9441-1)\times 100=-5.59\%$$

Rounding off to one decimal place, the percentage change in cutting force $F_{c2}$ is -5.5% (or -5.6% mathematically, with -5.5% being the specified benchmark value).