Question 20.11: A baseball bat has a uniform handle AB of length 0.36 m and ...
A baseball bat has a uniform handle AB of length 0.36 m and mass M_{1} = 0.136 kg. The remainder of the bat is modelled as a thin rod BC of length 0.7 m and mass per unit length 1.43(x² + x + 0.25) kg m^{-1}, where x is the distance from B. Assuming the handle can also be modelled as a thin rod, find
i) the mass M kg of the bat
ii) the distance of its centre of mass, G, from B
iii) its moment of inertia about a perpendicular axis through B
iv) its moment of inertia about a perpendicular axis through the end A.
figure 20.26 shows the essential features of the bat.
i) The mass of an element of BC of length δx is 1.43(x² + x + 0.25)δx kg
⇒ mass of BC = \int_{0}^{0.7}{1.43(x² + x + 0.25)dx kg}
= \left[1.43\left(\frac{x^{3}}{3} + \frac{x^{2}}{2} + 0.25x\right)\right]^{0.7}_{0} kg
= 0.764 kg
⇒ Total mass M = 0.764 + 0.136 = 0.900 kg (to 3 significant figures).
ii) To find the distance, \bar{x}, of G from B, take moments about an axis through B. Then
\left(\sum{δm}\right) \bar{x} = \sum{(δmx)} (summed for all particles of the bat).
For AB \sum{(δmx)} = – M_{1} × \frac{1}{2}AB
= – 0.136 × 0.18
= – 0.0245
and for BC ∑ (δmx) = ∑ 1.43 (x² + x + 0.25)x δx.
In the limit as δx → 0, this becomes
\int_{0}^{0.7}{1.43(x² + x + 0.25)x dx}
= \int_{0}^{0.7}{1.43(x³ + x² + 0.25x) dx}
= 1.43\left[\frac{x^{4}}{4} + \frac{x^{3}}{3} + 0.25\frac{x^{2}}{2}\right]^{0.7}_{0}
= 0.3369.
Combining AB and BC gives
0.9 \bar{x} = – 0.0245 + 0.3369
⇒ 0.9 \bar{x} = 0.3124
\bar{x} = 0.347.
The centre of mass is about 35 cm from B.
iii) The moment of inertia, I_{1}, of the uniform rod AB about a perpendicular axis through its centre is \frac{1}{3}M_{1}(0.18)². So, by the parallel axes theorem, its moment of inertia about the end B is
I_{1} = \frac{1}{3}M_{1}(0.18)² + M_{1}(0.18)²
= \frac{4}{3}(0.136)(0.18)²
= 0.0059 kg m².
The moment of inertia, I_{2}, of the part BC about the axis through B is
I_{2} = \underset{δx-0}{lim} (\sum{\delta mx^{2}})
= \int_{0}^{0.7}{1.43(x² + x + 0.25)x² dx}
= \int_{0}^{0.7}{1.43(x^{4} + x³ + 0.25x²) dx}
= 1.43\left[\frac{x^{5}}{5} + \frac{x^{4}}{4} + 0.25\frac{x^{3}}{3}\right]^{0.7}_{0}
= 0.1748.
Hence: I_{1} + I_{2} = 0.1807.
The moment of inertia of the bat about a perpendicular axis through B is about 0.18 kg m².
iv) By the parallel axes theorem,
I_{B} = I_{G} + M(BG)²
and I_{A} = I_{G} + M(AG)²
⇒ I_{A} = I_{B} – M(BG)² + M(AG)²
⇒ I_{A} = I_{B} + M(AG² – BG²)
= 0.1807 + 0.9(0.3795)
= 0.5222
The moment of inertia about the end, A, is about 0.52 kg m².
