3. The figure shows the total acceleration of a particde moving clockwise in a c

Question

3. The figure shows the total acceleration of a particde moving clockwise in a cirle of radlus at a certain instant of time, t. For that instant (ti), the magnitude of the total R-2m acceleration is a 10.0 m/s? and it makes an angle 25 relative to the radius. Consider the magnitudes of the tangential and angular accelerations of the particle to be constant R-2m a 10.0 925 t2 (5 Points) (a) the radial (centripetal) acceleration of the particle (5 Points) (b) the speed of the particle vy (5 Points) (c) the magnitude of the particle's angular acceleration, a.

Explanation / Answer

Q3.

part a:

centripetal acceleration=a*cos(theta)

=10*cos(25)=9.0631 m/s^2

part b:

as centripetal acceleration=speed^2/radius

==>v1^2/R=9.0631

==>v1=4.2575 m/s

part c:

angular acceleration=tangential acceleration/radius

=10*sin(25)/2

=2.1131 rad/s^2

part d:

at t1, angular velocity=v1/r=4.2575/2=2.12875 rad/s

from t1 to t2, it has completed an angle of 90 degrees=pi/2 rad

hence angular speed at t2^2-angular speed at t1^2=2*angular acceleration*ngle moved

==>angular speed at t2^2-2.12875^2=2*2.1131*pi/2

==>angular speed at t2=3.34216 rad/s

then v2=angular speed at t2*R

=6.68432 m/s

part e:

tangential acceleration remains constant at 10*sin(25)=4.22618 m/s^2

radial acceleration=v2^2/R=22.34 m/s^2

net magnitude of accleeration=sqrt(4.22618^2+22.34^2)

=22.736 m/s^2

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