Question

`AB` is a horizontal diameter of a ball of mass `m=0.4 kg` and radius `R=0.10m`. At time `t=0`, a sharp impulse is applied at `B` at angle of `45^@` with the horizontal as shown in figure so that the ball immediately starts to move with velocity `v_(0)=10ms^(-1)`

a. Calculate the impulse and angular velcity of ball just after impulse provided.
If coefficient of kinetic friction between the floor and th ball is `mu=0.1`. Calculate. The
b. velocity of ball when it stops sliding,
c. time t at that instant
d. horizontal distance travelled by the ball up to that instant.
e. angular displacement of the ball about horizontal diameter perpendicular to `AB`, up to that instant, and
f. energy lost due to friction (`g=10ms^(-2))`

Answer 1

Since, line of action of impulse does not pass through centre of mass of the sphere, therefore, just after application of impulse, the sphere starts to move, both translationally and rotationally. Translation motion is produced by horizontal component of the impulse, while rotational motion is produced by momnt of the impulse. Let the impulse applied be `J`.
Then its horizontal coponent provides horizontal motion.
`J.cos45^@` which gives `J=4sqrt(2)kgms^(-1)`
Moment of inertial of ball about centroidal axis is
`I=2/4mR^(2)=1.6xx10^(-30kgm^(2)`
Initial angular moment of ball (about centre) =`J(Rsin45^@)`
Angular impule will change angular momentum of the ball.
`J.R sin 45^@=/_L=(Iomega_(0)-0)`
Which gives `omega_(0)=250rads^(-1)`(clockwise)