Question

(a) What is the tangential acceleration of a bug on the rim of a 13.0-in.-diameter disk if the disk accelerates uniformly from rest to an angular speed of 80.0 rev/min in 4.30s?_____ m/s2
(b) When the disk is at its final speed, what is the magnitude of the tangential velocity of the bug? m/s
(c) One second after the bug starts from rest, what is the magnitude of its tangential acceleration? m/s2
(d) One second after the bug starts from rest, what is the magnitude of its centripetal acceleration? m/s2
(e) One second after the bug starts from rest, what is its total acceleration? (Take the positive direction to be in the direction of motion.) magnitude m/s2 direction ° from the radially inward direction

Answer 1

Answer:

0.321659377 m/s²

1.383138458 m/s

0.321659377 m/s²

0.62667 m/s²

0.7044 m/s² and 27.17°

Explanation:

d = Diameter of rim = 13 in = $13\times 0.0254=0.3302\ m$

r = Radius = $\frac{d}{2}=\frac{0.3302}{2}=0.1651\ m$

$\omega_f$ = Final angular velocity = $80\times\frac{2\pi}{60}=8.37758\ rad/s$

$\omega_i$ = Initial angular velocity = 0

$\alpha$ = Angular acceleration

t = Time taken = 4.3 s

Equation of rotational motion

$\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{8.37758-0}{4.3}\\\Rightarrow \alpha=1.94827\ rad/s^2$

Tangential acceleration is given by

$a_t=r\alpha\\\Rightarrow a_t=0.1651\times 1.94827\\\Rightarrow a_t=0.321659377\ m/s^2$

The tangential acceleration of the bug is 0.321659377 m/s²

Tangential velocity is given by

$v=r\omega\\\Rightarrow v=0.1651\times 8.37758\\\Rightarrow v=1.383138458\ m/s$

The tangential velocity of the bug is 1.383138458 m/s

The tangential acceleration is constant which is 0.321659377 m/s²

Centripetal acceleration is given by

$a_c=\frac{a_tt^2}{r}\\\Rightarrow a_c=\frac{0.321659377^2\times 1}{0.1651}\\\Rightarrow a_c=0.62667\ m/s^2$

The centripetal acceleration of the bug is 0.62667 m/s²

The resultant of the acceleration gives us total acceleration

$a=\sqrt{a_t^2+a_c^2}\\\Rightarrow a=\sqrt{0.321659377^2+0.62667^2}\\\Rightarrow a=0.7044\ m/s^2$

Direction is given by

$\theta=cos^{-1}\frac{a_c}{a}\\\Rightarrow \theta=cos^{-1}\frac{0.62667}{0.7044}\\\Rightarrow \theta=27.17^{\circ}$

The magnitude and direction of the acceleration is 0.7044 m/s² and 27.17°