Question

A discus thrower (with arm length of \(1.20 \mathrm{~m}\) ) starts from rest and begins to rotate counterclockwise with an angular acceleration of \(2.50 \mathrm{rad} / \mathrm{s}^{2}\) a) How long does it take the discus thrower's speed to get to \(4.70 \mathrm{rad} / \mathrm{s} ?\) b) How many revolutions does the thrower make to reach the speed of \(4.70 \mathrm{rad} / \mathrm{s} ?\) c) What is the linear speed of the discus at \(4.70 \mathrm{rad} / \mathrm{s}\) ? d) What is the linear acceleration of the discus thrower at this point? e) What is the magnitude of the centripetal acceleration of the discus thrown? f) What is the magnitude of the discus's total acceleration?

Short answer

Answer: The magnitude of the discus's total acceleration after reaching the angular speed of 4.70 rad/s is 26.63 m/s².

Step-by-step solution

Find time to reach the angular speed of \(4.70\ \mathrm{rad/s}\)

First, we define the following variables: \(\alpha = 2.50\ \mathrm{rad/s^2}\) is the angular acceleration, and \(\omega = 4.70\ \mathrm{rad/s}\) is the angular speed we want to find the time for. We can use the following equation which relates the two variables with time: $$\omega = \alpha t$$ Now, we can solve for the time (t): $$t = \frac{\omega}{\alpha} = \frac{4.70\ \mathrm{rad/s}}{2.50\ \mathrm{rad/s^2}} = 1.88\ \mathrm{s}$$ #b)

Find the number of revolutions made to reach the angular speed

To find the number of revolutions made, we need to find the total angle, \(\Theta\), while the discus thrower reaches the angular speed of \(4.70 \mathrm{rad/s}\). We can use the following angular equation: $$\Theta = \frac{1}{2} \alpha t^2$$ Substituting the given values: $$\Theta = \frac{1}{2} \times 2.50\ \mathrm{rad/s^2} \times (1.88\ \mathrm{s})^2 = 4.42\ \mathrm{rad}$$ Now, we need to convert radians to revolutions, by dividing by \(2\pi\): $$\text{Revolutions}=\frac{\Theta}{2\pi} = \frac{4.42\ \mathrm{rad}}{2\pi}\approx 0.704\ \text{revolutions}$$ #c)

Find the linear speed at \(4.70\ \mathrm{rad/s}\)

To find the linear speed (v) at the given angular speed, we first define the arm's length as \(r = 1.20\ \mathrm{m}\). Then, we use the relationship between linear and angular speed: $$v = r\omega$$ Substituting the values: $$v = 1.20\ \mathrm{m} \times 4.70\ \mathrm{rad/s} = 5.64\ \mathrm{m/s}$$ #d)

Find the linear acceleration at this point

To find the linear acceleration (a) at this point, we need to use the relationship between linear and angular acceleration: $$a = r\alpha$$ Substituting the values: $$a = 1.20\ \mathrm{m} \times 2.50\ \mathrm{rad/s^2} = 3.00\ \mathrm{m/s^2}$$ #e)

Find the magnitude of the centripetal acceleration

To find the centripetal acceleration \((a_c)\), we use the following formula: $$a_{c}=\omega^2r$$ Plugging in the values: $$a_{c}=(4.70\ \mathrm{rad/s})^2 \times 1.20\ \mathrm{m}=26.37\ \mathrm{m/s^2}$$ #f)

Find the magnitude of the discus's total acceleration

To find the total acceleration \((a_T)\), we use the Pythagorean theorem, as the linear acceleration and centripetal acceleration are perpendicular to each other: $$a_T=\sqrt{a^2+a_c^2}$$ Substituting the values: $$a_T= \sqrt{(3.00\ \mathrm{m/s^2})^2+(26.37\ \mathrm{m/s^2})^2}=26.63\ \mathrm{m/s^2}$$