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 magnitude of the tangential speed of an object placed 10.0 cm from the center of the wheel is A) 1.0 m/s. B) 0.50 m/s. C) 0.0 m/s. D) 1.5 m/s. E) 2.0 m/s.

Question

Quizplus · Accepted Answer

The tangential speed $v$ of an object at a distance $r$ from the center of rotation is given by $v = r\omega$, where $\omega$ is the angular velocity of the object. The angular velocity $\omega$ can be found by differentiating the given equation for $\phi(t)$ with respect to time $t$, which gives $\omega = \frac{d\phi}{dt} = c + 3dt^2$. Substituting $c = 1.0 \, \text{rad/s}$, $d = -0.50 \, \text{rad/s}^3$, and $t = 2.0 \, \text{s}$ into this equation yields $\omega = 1.0 + 3(-0.50)(2.0)^2 = 1.0 - 6.0 = -5.0 \, \text{rad/s}$. The magnitude of $\omega$ is $5.0 \, \text{rad/s}$. Then, with $r = 10.0 \, \text{cm} = 0.1 \, \text{m}$, the tangential speed is $v = 0.1 \times 5.0 = 0.5 \, \text{m/s}$.

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 $\phi ( \mathrm { rad } ) = c t + d t ^ { 3 }$