The graph below gives the position x vs. t for a spot on rim of a rotating wheel

Question

The graph below gives the positionxvs.tfor a spot on rim of a rotating wheel.

From the graph determine the following quantities:

(a) The angular speed?

- (1.57 rad/s); (2.50 rad/s); (14.1 rad/s); (24.7 rad/s); (.0637 rad/s); (15.7 m/s) ;

(15.7 rad/s) ;(24.7 m/s) ;(.100 rad/s) ;(7.5 rad/s) ; (17.3 rad/s) ; (0 rad/s)

(b) The speed of the spotv:

- (1.57 rad/s); (2.50 rad/s); (14.1 rad/s); (24.7 rad/s); (.0637 rad/s); (15.7 m/s) ;

(15.7 rad/s) ;(24.7 m/s) ;(.100 rad/s) ;(7.5 rad/s) ; (17.3 rad/s) ; (0 rad/s)

(c) The magnitude of centripetal acceleration of the spotac

-(0 m/s) ; ( 24.7 m/s^2) ; (0 m/s^2) ; (15.7 rad/s^2) (15.7 m/s^2) (7.85 rad/s^2(

(21.0 m/s^2) (.250 m/s) (26.6 m/s^2)

(d) The magnitude of angular acceleration?:

- (21.5 m/s^2) (1.57 m/s^2) (0 rad/s^2) (24.7 rad/s^2)

(27.6 m/s^2) (24.7 m/s^2) ( 0 m/s^2) (39.3 rad/s^2)

(e) The magnitude of tangential acceleration of the spotat:

(24.7 rad/s^2) (22.7 m/s^2) (1.57 m/s^2) (0 m/s^2)(24.7 m/s^2)

(21.5 m/s^2) (30.8 m/s^2)(0 rad/s^2) (27.9 m/s^2)

Explanation / Answer

(a)from 10 to 10 again means 2pi radians in 0.4 s
so =2pi/0.4=15.7 rad/s

(b)speed=r=-0.1*15.7=-1.57 m/s

(c)a=r^2=0.1*15.7^2=24.7 m/s^2

(d)=0 rad/s^2

as is constant

(e)a=r=0

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