Answer:
[tex]\tau=22.13Nm[/tex]
Explanation:
information we have:
mass: [tex]m=0.9kg[/tex]
lenght: [tex]L=0.95m[/tex]
frequency: [tex]f=2.6rev/s[/tex]
time: [tex]t=0.2s[/tex]
and from the information we have we can calculate the angular velocity [tex]\omega[/tex]. which is defined as
[tex]\omega=2\pi f[/tex]
[tex]\omega=2\pi (2.6rev/s)\\\omega=16.336 rev/s[/tex]
----------------------------
Now, to calculate the torque
We use the formula
[tex]\tau=I \alpha[/tex]
where [tex]I[/tex] is the moment of inertia and [tex]\alpha[/tex] is the angular acceleration
moment of inertia of a uniform rod about the end of it:
[tex]I=\frac{1}{3}mL^2[/tex]
substituting known values:
[tex]I=\frac{1}{3} (0.9kg)(0.95m)^2\\I=0.271kg/m^2[/tex]
for the torque we also need the acceleration [tex]\alpha[/tex] which is defined as:
[tex]\alpha=\frac{\omega}{t}[/tex]
susbtituting known values:
[tex]\alpha=\frac{16.336rev/s}{0.2s} \\\alpha=81.68rev/s^2[/tex]
and finally we substitute [tex]I[/tex] and [tex]\alpha[/tex] into the torque equation [tex]\tau=I \alpha[/tex]:[tex]\tau=(0.271kg/m^2)(81.68rev(s^2)\\\tau=22.13Nm[/tex]
The torque the player applies is mathematically given as
t=22.13Nm
Question Parameter(s):
a bat, accelerating it from rest to 2.6 rev/srev/s in a time of 0.20 s
the bat has a 0.90-kg mass
Generally, the equation for the Torque is mathematically given as
T=Ia
Where
I=1/3mL^2
I=1/3(0.9kg)(0.95m)^2
I=0.271kg/m^2
And
[tex]a=\frac{16.336rev/s}{0.2s}[/tex]
a=81.68rev/s^2
In conclusion, The torque
t=(0.271)(81.68)
t=22.13Nm
Read more about Torque
https://brainly.com/question/20691242
