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Andrews [41]
3 years ago
12

A baseball bat could be considered a uniform rod. It ha a mass of 0.97 kg and a length of 97 cm. If a player accelerates it from

rest to 2.8 rev/s in 0.20s, how much torque is applied?
Physics
2 answers:
grigory [225]3 years ago
5 0

Answer:

The torque is 26.75 N-m.

Explanation:

Given that,

Mass = 0.97 kg

Length = 97 cm

Time = 0.20 s

Angular speed \omega= 2.8 rev/s = 2.8\times2\pi\ rad/s

We need to calculate the moment of inertia of rod at one end

Using formula of the moment of inertia

I=\dfrac{Ml^2}{3}

Put the value into the formula

I=\dfrac{0.97\times(0.97\times10^{-2})^2}{3}

I=0.3042\ kg m^2

We need to calculate the angular acceleration

Using formula of angular acceleration

\alpha=\dfrac{\Delta \omega}{\Delta t}

Put the value into the formula

\alpha=\dfrac{2.8\times2\pi}{0.20}

\alpha=87.964\ rad/s^2

We need to calculate the torque

Using formula of torque

\tau=I\times\alpha

Put the value into the formula

\tau=0.3042\times87.964

\tau=26.75\ N-m

Hence, The torque is 26.75 N-m.

sukhopar [10]3 years ago
4 0

Answer:

Applied torque, T = 26.74 N-m

Given:

Mass of rod, M = 0.97 kg

Length of the rod, L = 97 cm = 0.97 m

Change in angular velocity, \Delta \omega = 2.8 rev/s = 2.8\times 2\pi = 17.59 rad/s

time, t = 0.20 s

Solution:

Now, we know that torque is given by:

T = I\alpha                          (1)

where

I = moment of inertia of the rod

\alpha  = angular acceleration

Using perpendicular axis formula to calculate the moment of inertia:

I = \frac{ML^{2}}{12}                    

Now, the base ball bat will be held from one end, therefore,  the moment of inertia about the axis at the end is given by:

I = \frac{ML^{2}}{12} + \frac{ML^{2}}{4} = \frac{ML^{2}}{3}

I = \frac{0.97\times (0.97)^{2}}{3} = 0.304 kg-m^{2}

Now, angular acceleration can be calculated as:

\alpha = \frac{\Delta \omega}{t} = \frac{17.59}{0.20} = 87.95 rad/s^{2}

Now, calculation of torque using eqn (1):

T = 0.304\times 87.95 = 26.74 N-m

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At point A

The volume of the gas, V₁ = 5.00 L

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The temperature of the gas, T₁ = 300 K

At point B

The volume of the gas, V₂ = V₁ = 5.00 L

The pressure of the gas, P₂ = 3.00 atm

The temperature of the gas, T₂ = Not given

At point C

The volume of the gas, V₃ = Not given

The pressure of the gas, P₃ = 1 atm

The temperature of the gas, T₂ = T₃ = 300 K

(a) The ideal gas equation is given as follows;

P·V = n·R·T

Where;

P = The pressure of the gas

V = The volume of the gas

n = The number of moles present

R = The universal gas constant = 0.08205 L·atm·mol⁻¹·K⁻¹

n = PV/(R·T)

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P₁/T₁ = P₂/T₂

∴ T₂ = P₂·T₁/P₁

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T₂ = 3.0 atm. × 300 K/(1.00 atm.) = 900 K

The temperature at point B, T₂ = 900 K

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(d) For a constant temperature process, according to Boyle's law, we have;

P₂·V₂ = P₃·V₃

V₃ = P₂·V₂/P₃

∴ V₃ = 3.00 atm. × 5.00 L/(1.00 atm.) = 15 L

The volume at point C, V₃ = 15 L

(e) The process A → B, which is a constant volume process, can be carried out in a vessel with a fixed volume

The process B → C, which is a constant temperature process, can be carried out in an insulated adjustable vessel

The process C → A, which is a constant pressure process, can be carried out in an adjustable vessel with a fixed amount of force applied to the piston

(f) For A → B, W = 0,

Q = Eint = n·cv·(T₂ - T₁)

Cv for monoatomic gas = 3/2·R

∴ Q = 0.203 moles × 3/2×0.08205 L·atm·mol⁻¹·K⁻¹×(900 K - 300 K) = 1,518.91596 J

Q = Eint = 1,518.91596 J

For B → C, we have a constant temperature process

Q = n·R·T₂·㏑(V₃/V₂)

∴ Q = 0.203 moles × 0.08205 L·atm/(mol·K) × 900 K × ln(15 L/5.00 L) ≈ 1668.69974 J

Eint = 0

Q = W ≈ 1668.69974 J

For C → A, we have a constant pressure process

Q = n·Cp·(T₁ - T₃)

∴ Q = 0.203 moles × (5/2) × 0.08205 L·atm/(mol·K) × (300 K - 900 K) = -2,531.5266 J

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∴ W = 1.00 atm × (5.00 L - 15.00 L) = -1,013.25 J

W = -1,013.25 J

Eint = n·Cv·(T₁ - T₃)

Eint = 0.203 moles × (3/2) × 0.08205 L·atm/(mol·K) × (300 K - 900 K) = -1,518.91596 J

Eint = -1,518.91596 J

(g) ∑Q = 1,518.91596 J + 1668.69974 J - 2,531.5266 J = 656.089 J

∑W = 0 + 1668.69974 J -1,013.25 J = 655.449 J

∑Eint = 1,518.91596 J + 0 -1,518.91596 J = 0 J

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