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Chapter 5: Problem 556

A car is moving at a speed of \(72 \mathrm{~km} / \mathrm{hr}\) the radius of its wheel is \(0.25 \mathrm{~m}\). If the wheels are stopped in 20 rotations after applying breaks then angular retardation produced by the breaks is \(\ldots .\) \(\\{\mathrm{A}\\}-25.5 \mathrm{rad} / \mathrm{s}^{2}\) \(\\{\mathrm{B}\\}-29.52 \mathrm{rad} / \mathrm{s}^{2}\) \(\\{\mathrm{C}\\}-33.52 \mathrm{rad} / \mathrm{s}^{2}\) \(\\{\mathrm{D}\\}-45.52 \mathrm{rad} / \mathrm{s}^{2}\)

Short Answer

Expert verified
The angular retardation (α) produced by the brakes is -29.52 \(rad/s^2\). So, the correct answer is: \(\\{\mathrm{B}\\}-29.52 \mathrm{rad} / \mathrm{s}^{2}\)

Step by step solution

01

Converting linear speed to angular speed

We need to convert the linear speed (v) to the initial angular speed (ω₁). Use the formula: \(v = ω × r\) where v is the linear speed, ω is the angular speed, and r is the radius of the wheel. Given values: \(v = 72 \frac{km}{hr}\) \(r = 0.25 m\) Now, convert the linear speed to meters per second: \(v = 72 \frac{km}{hr} × \frac{1,000 m}{1 km} × \frac{1 hr}{3,600 s} = 20 m/s\) Using the formula: \(ω_{1} = \frac{v}{r}\)
02

Calculate initial angular velocity (ω₁)

Plug the values into the formula for ω₁: \(ω_{1} = \frac{20}{0.25} = 80 rad/s\)
03

Calculate final angular velocity (ω₂)

If the wheels are stopped in 20 rotations (meaning the final angular velocity ω₂ = 0), we have: \(ω_{2} = 0 rad/s\)
04

Calculate angular displacement (θ)

Since there are 20 rotations, we need to convert rotations to radians. The angular displacement (θ) is the total angle covered during these 20 rotations. Recall that 1 rotation = \(2\pi\) radians, so: \(θ = 20 × 2\pi = 40\pi rad\)
05

Find the angular retardation (α)

Now, use the relationship between angular velocities, angular displacement, and angular acceleration: \(ω_{2}^2 = ω_{1}^2 + 2αθ\) Plug in ω₁, ω₂, and θ to solve for α: \(0 = 80^2 + 2α(40\pi)\) Solving for α, we get: \(α = -\frac{80^2}{80\pi} = -29.52 rad/s^2\) The angular retardation (α) produced by the brakes is -29.52 \(rad/s^2\). So, the correct answer is: \(\\{\mathrm{B}\\}-29.52 \mathrm{rad} / \mathrm{s}^{2}\)

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