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Automobile Engine The rotational speed of an automobile engine is increased at a constant rate from 1200 rev/min to 3000 rev \(/ \mathrm{min}\) in \(12 \mathrm{~s}\). (a) What is its rotational acceleration in revolutions per minute-squared? (b) How many revolutions does the engine make during this 12 s interval?

Short Answer

Expert verified
a) 9000 rev/min^2. b) 420 revolutions.

Step by step solution

01

Convert Time to Minutes

Convert the given time from seconds to minutes to be consistent with the units of revolutions per minute. Since there are 60 seconds in a minute, \[ t = \frac{12 \text{ s}}{60 \text{ s/min}} = 0.2 \text{ min} \]
02

Determine the Change in Rotational Speed

Calculate the change in rotational speed (Δω) in revolutions per minute (rev/min). The initial speed (ω_i) is 1200 rev/min and the final speed (ω_f) is 3000 rev/min. \[ \text{Δω} = 3000 \text{ rev/min} - 1200 \text{ rev/min} = 1800 \text{ rev/min} \]
03

Calculate Rotational Acceleration

Rotational acceleration (α) is the change in rotational speed divided by the time interval. The time interval is 0.2 minutes (from Step 1). \[ α = \frac{\text{Δω}}{t} = \frac{1800 \text{ rev/min}}{0.2 \text{ min}} = 9000 \text{ rev/min}^2 \]
04

Use the Rotational Kinematic Equation

Determine how many revolutions the engine makes using the rotational kinematic equation: \[ \theta = ω_i t + \frac{1}{2} α t^2 \] where θ is the angular displacement in revolutions, ω_i is the initial rotational speed, α is the rotational acceleration, and t is the time.
05

Plug in Known Values

Substitute the known values into the equation to find θ. \[ θ = (1200 \text{ rev/min} \times 0.2 \text{ min}) + \frac{1}{2} (9000 \text{ rev/min}^2 \times (0.2 \text{ min})^2) \] Simplify the expression: \[ θ = 240 \text{ rev} + \frac{1}{2} (9000 \text{ rev/min}^2 \times 0.04 \text{ min}^2) \] \[ θ = 240 \text{ rev} + \frac{1}{2} (360 \text{ rev}) \] \[ θ = 240 \text{ rev} + 180 \text{ rev} = 420 \text{ rev} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rotational Acceleration
Rotational acceleration is an important concept in rotational kinematics. Think of it as how much the speed of a rotating object changes over time. For instance, if an automobile engine speeds up from 1200 revolutions per minute (RPM) to 3000 RPM in a certain time, the change in speed is what we're interested in.
We use the formula \( \alpha = \frac{\Delta \omega}{t} \) to find the rotational acceleration \( \alpha \). Here, \( \Delta \omega \) is the change in rotational speed, and \( t \) is the time over which this change occurs.
In our example, the engine's speed changes by 1800 RPM in 0.2 minutes. Plugging these values into the formula gives: \[ \alpha = \frac{1800 \text{ rev/min}}{0.2 \text{ min}} = 9000 \text{ rev/min}^2 \]
This tells us how quickly the engine is speeding up.
Angular Displacement
Angular displacement, often denoted as \( \theta \), represents the angle through which an object rotates. It's like figuring out the distance a car travels, but for rotational movement.
To find \( \theta \), we use the rotational kinematic equation: \[ \theta = \omega_i t + \frac{1}{2} \alpha t^2 \]
In this equation, \( \omega_i \) is the initial rotational speed, \( t \) is the time, and \( \alpha \) is the rotational acceleration.
For the engine example: \( \omega_i = 1200 \text{ rev/min} \), \( t = 0.2 \text{ min} \), and \( \alpha = 9000 \text{ rev/min}^2 \).
Plugging in these values: \[ \theta = (1200 \text{ rev/min} \times 0.2 \text{ min}) + \frac{1}{2} (9000 \text{ rev/min}^2 \times (0.2 \text{ min})^2) \]
Simplifying, we get: \[ \theta = 240 \text{ rev} + \frac{1}{2} (360 \text{ rev}) = 240 \text{ rev} + 180 \text{ rev} = 420 \text{ rev} \]
This means the engine makes 420 revolutions during the time interval.
Revolutions per Minute
Revolutions per minute (RPM) is a unit that measures rotational speed. It's like the speedometer in your car but for how fast something spins.
RPM tells you how many complete turns an object makes in one minute. If an engine has a speed of 1200 RPM, it means the engine completes 1200 full rotations every minute.
When solving problems, it's crucial to keep track of these units. For example, converting time to minutes helps maintain consistency with RPM.
In the original exercise, the engine's speed increased from 1200 RPM to 3000 RPM. This change or increase is a key aspect of rotational motion and helps us determine the rotational acceleration and angular displacement.
By understanding and correctly using RPM, you can better analyze and solve rotational kinematics problems.

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Most popular questions from this chapter

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