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babunello [35]
3 years ago
10

A softball player swings a bat, accelerating it from rest to 2.6 rev/srev/s in a time of 0.20 ss . Approximate the bat as a 0.90

-kgkg uniform rod of length 0.95 mm, and compute the torque the player applies to one end of it.
Physics
1 answer:
svetoff [14.1K]3 years ago
5 0

Answer:

\tau=22.13Nm

Explanation:

information we have:

mass: m=0.9kg

lenght: L=0.95m

frequency: f=2.6rev/s

time: t=0.2s

and from the information we have we can calculate the angular velocity \omega. which is defined as

\omega=2\pi f

\omega=2\pi (2.6rev/s)\\\omega=16.336 rev/s

----------------------------

Now, to calculate the torque

We use the formula

\tau=I \alpha

where I  is the moment of inertia and \alpha is the angular acceleration

moment of inertia of a uniform rod about the end of it:

I=\frac{1}{3}mL^2

substituting known values:

I=\frac{1}{3} (0.9kg)(0.95m)^2\\I=0.271kg/m^2

for the torque we also need the acceleration \alpha which is defined as:

\alpha=\frac{\omega}{t}

susbtituting known values:

\alpha=\frac{16.336rev/s}{0.2s} \\\alpha=81.68rev/s^2

and finally we substitute I and  \alpha  into the torque equation \tau=I \alpha:\tau=(0.271kg/m^2)(81.68rev(s^2)\\\tau=22.13Nm

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Answer:

θ  = 12.95º

Explanation:

For this exercise it is best to separate the process into two parts, one where they collide and another where the system moves altar the maximum height

Let's start by finding the speed of the bar plus clay ball system, using amount of momentum

The mass of the bar (M = 0.080 kg) and the mass of the clay ball (m = 0.015 kg) with speed (v₀ = 2.0 m / s)

Initial before the crash

      p₀ = m v₀

Final after the crash before starting the movement

     p_{f} = (m + M) v

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Lower

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Higher

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   Em₀ = E_{mf}

   ½ (m + M) v² = (m + M) g y

   y = ½ v² / g

   y = ½ 0.316² / 9.8

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Let's look for the angle the height from the pivot point is

    L = 0.40 / 2 = 0.20 cm

The distance that went up is

     y = L - L cos θ

     cos θ  = (L-y) / L

     θ  = cos⁻¹ (L-y) / L

     θ  = cos⁻¹-1 ((0.20 - 0.00509) /0.20)

      θ  = 12.95º

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