a) The angular speed of point B is equal to the angular speed of point A: [tex]\omega_B = \omega_A[/tex]
b) The linear speed of point B is twice the linear speed of point A: [tex]v_B =2 v_A[/tex]
Explanation:
a)
The angular velocity of a point in circular motion is given by
[tex]\omega = \frac{\Delta \theta}{\Delta t}[/tex]
where
[tex]\Delta \theta[/tex] is the angular displacement
[tex]\Delta t[/tex] is the time interval considered
Here we have two points on a rotating wheel: point A and point B. Point B is twice as far as point A from the axis. However, all the points of a uniform disk (such as in this case) covers the same angle in the same time interval. This means that the angular velocity of all the points of a rotating wheel is the same: therefore, the angular speed of point B is equal to the angular speed of point A,
[tex]\omega_B = \omega_A[/tex]
b)
The linear speed of a point on the disk is given by
[tex]v=\omega r[/tex]
where
[tex]\omega[/tex] is the angular speed
r is the distance of the point from the axis of rotation
For point A, we have
[tex]v_A = \omega_A r_A[/tex]
For point B, we have
[tex]v_B = \omega_B r_B[/tex]
However, we said that the angular speed of the two points is the same:
[tex]\omega_B = \omega_A[/tex]
But point B is twice as far as point A from the axis:
[tex]r_B = 2 r_A[/tex]
Therefore, we get
[tex]v_B = \omega_A (2r_A) = 2(\omega_A r_A) = 2 v_A[/tex]
So, the linear speed of point B is twice the linear speed of point A.
Learn more about angular speed:
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