Question

The velocity ofa particle is defined by v (-2i+(3-20A, m/s where t is in seconds. If r 0 when t 0, determine the displacement of the particle during the time interval t As (-8/-4A, m As (-6i+-6A. m As (-10/-2A.m As-(-4, m 1 s and t 4 s. A motor gives a disk A an angular acceleration of aA (2t+3) rads, where t is in seconds. If the inidal angular velocity of the disk is Dg 5 rads, determine the magnitude of the velocity of biock B when t 2 s va-1.5 m/s v- 15 m/s ve-7 m/s ve-2.25 m/s The angular velocity of the disk shown in the figure is defined by (5+20 ad/s, where tis in seconds. Determine the magnitude of the velocity of point A on the disk when t 0.5 s. esa VA 2.25 m/s VA 1.8 m/s VA-7 m/s VA 5.6 m/s A Moving to another question will save this response.

Answer 1

a) In the first part (image), we know that velocity v= ds/dt where s= displacement. Now, $\Delta$s =
vdt
.

so integrating velocity with respect to t from t= 1sec to t= 4sec, we get $\Delta$s= (-6i + (-6j) ) m

b) For the second part, we know that angular acceleration $\alpha$ = d$\omega$/dt where $\omega$ = angular velocity. Now .

adt

we know angular velocity initially at t = 0 sec is 5 rad/s from this we get $\omega_{2}$ = 15 rad/s. We also know v= r x $\omega$ and this implies velocity of block b, since it is attached at the end of the disk, tangentially and perpendicular to r, we get velocity to be 2.25rad/s.

c) With the same fundamental as above, v= r x $\omega$, r= 0.8 m, $\omega$ (at t= 0.5s) = 2.25 rad/s, velocity at A is 1.8 m/s.