A 42.0-cm diameter disk rotates with a constant angular acceleration of 2.70 rad/s2. It starts from rest at t = 0, and a line drawn from the center of the disk to a point P on the rim of the disk makes an angle of 57.3° with the positive x-axis at this time.

(a) Find the angular speed of the wheel at t=2.30s.

(b) Find the linear velocity and tangential acceleration of P at t=2.30s.

(c) Find the position of P (in degrees, with respect to the positive x-axis) at t = 2.30s.

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boffeemadrid boffeemadrid · Answer 1 · 2020-01-10T07:05:05+01:00

Answer:

6.21 rad/s

1.3041 m/s, 0.567 m/s²

[tex]106.4778\ ^{\circ}[/tex]

Explanation:

[tex]\omega_f[/tex] = Final angular velocity

[tex]\omega_i[/tex] = Initial angular velocity = 0

[tex]\alpha[/tex] = Angular acceleration = 2.3 rad/s²

[tex]\theta[/tex] = Angle of rotation

t = Time taken = 2.3 s

Equation of rotational motion

[tex]\omega_f=\omega_i+\alpha t\\\Rightarrow \omega_f=0+2.7\times 2.3\\\Rightarrow \omega_f=6.21\ rad/s[/tex]

The angular speed is 6.21 rad/s

Linear velocity is given by

[tex]v=r\omega\\\Rightarrow v=0.21\times 6.21\\\Rightarrow v=1.3041\ m/s[/tex]

Linear velocity is 1.3041 m/s

Tangential acceleration is given by

[tex]a_t=r\alpha\\\Rightarrow a_t=0.21\times 2.7\\\Rightarrow a_t=0.567\ m/s^2[/tex]

Tangential acceleration is 0.567 m/s²

[tex]\theta=\omega_it+\dfrac{1}{2}\alpha t^2\\\Rightarrow \theta=0\times t+\dfrac{1}{2}\times 2.7\times 2.3^2\\\Rightarrow \theta=7.1415\ rad[/tex]

In degress the angle would be

[tex]57.3+7.1415\times \dfrac{180}{\pi}=466.47780\ ^{\circ}[/tex]

From x axis it would be

[tex]466.47780-360=106.4778\ ^{\circ}[/tex]

The angle is [tex]106.4778\ ^{\circ}[/tex] from x axis