A wheel undergoes rotational motion according to the equation $\phi ( \mathrm { rad } ) = c t + d t ^ { 3 }$ , where $\surd$ is in radians, c = 1.0 rad/s, and d = -0.50 rad/s³. At t = 2.0 s the net acceleration of an object 10.0 cm from the center of the wheel is A) 0 m/s². B) 1.3 m/s². C) 2.6 m/s². D) 3.9 m/s².

Question

Quizplus · Accepted Answer

The net acceleration of an object in rotational motion consists of tangential acceleration ($a_t$) and centripetal acceleration ($a_c$). The tangential acceleration can be found by differentiating the given equation with respect to time to get angular acceleration ($\alpha$), and then multiplying by the radius ($r$). The centripetal acceleration is given by $a_c = r\omega^2$, where $\omega$ is the angular velocity, which can be found by differentiating the given equation once with respect to time.Given: $c = 1.0 \, \text{rad/s}$, $d = -0.50 \, \text{rad/s}^3$, and $r = 10.0 \, \text{cm} = 0.1 \, \text{m}$.1. Differentiate $\phi(t) = ct + dt^3$ to find $\omega(t) = \frac{d\phi}{dt} = c + 3dt^2$. At $t = 2.0 \, \text{s}$, $\omega = 1 + 3(-0.50)(2)^2 = 1 - 6 = -5 \, \text{rad/s}$.2. Differentiate $\omega(t)$ to find $\alpha(t) = \frac{d\omega}{dt} = 6dt$. At $t = 2.0 \, \text{s}$, $\alpha = 6(-0.50)(2) = -6 \, \text{rad/s}^2$.3. Calculate $a_t = \alpha r = -6 \times 0.1 = -0.6 \, \text{m/s}^2$.4. Calculate $a_c = r\omega^2 = 0.1 \times (-5)^2 = 0.1 \times 25 = 2.5 \, \text{m/s}^2$.5. The net acceleration is the vector sum of $a_t$ and $a_c$, but since they are perpendicular, we calculate the magnitude as $a = \sqrt{a_t^2 + a_c^2}$. Substituting the values gives $a = \sqrt{(-0.6)^2 + (2.5)^2} \approx \sqrt{0.36 + 6.25} = \sqrt{6.61} \approx 2.57 \, \text{m/s}^2$, which rounds to $2.6 \, \text{m/s}^2$.

A Wheel Undergoes Rotational Motion According to the Equation ϕ(rad)=ct+dt3\phi ( \mathrm { rad } ) = c t + d t ^ { 3 }ϕ(rad)=ct+dt3

A Wheel Undergoes Rotational Motion According to the Equation $\phi ( \mathrm { rad } ) = c t + d t ^ { 3 }$