Answer:
Explanation:
Moment of inertia of wheel = 1/2 x mR² , m is mass and R is radius of wheel
= .5 x 9 x .4²
= .72 kg m²
Torque created on wheel by string = T x r , T is tension and r is radius of wheel .
13 x .4 = 5.2 N m
angular acceleration α = torque / moment of inertia
= 5.2 / .72
= 7.222 rad /s²
a ) final angular speed = α x t , α is angular acceleration , t is time.
= 7.222 x .72
= 5.2 rad /s
b )
θ = 1/2 α t² , θ is angle turned , t is time
= .5 x 7.222 x .72²
= 1.872 rad
average angular speed = θ / t
= 1.872 / .72
= 2.6 rad /s
c )
angle turned = 1.872 rad ( discussed above )
d )
length of string coming off
= angle rotated x radius of wheel
= 1.872 x .4
= .7488 m .
74.88 cm
Explanation:
The position vector r:
The velocity vector v:
The acceleration vector a:
We really can't tell from the given information.
We don't know HOW MUCH Marv enlarged his cannonballs,
or HOW MUCH faster Seymour's balls became.
If we assume that they both, let's say, DOUBLED something,
then Seymour accomplished more, and the destructive capability
of his balls has increased more.
I say that because the destructive capability of a cannonball is
pretty much just its kinetic energy when it arrives and hits the target.
Now, we all know the equation for kinetic energy.
K.E. = (1/2) (mass) (speed-SQUARED) .
We can see right away that if Marv started shooting balls with
double the mass but at the same speed, then they have double
the kinetic energy of the old ones.
But if Seymour started shooting the same balls with double the SPEED,
then they have (2-SQUARED) as much kinetic energy as they used to.
That's 4 times as much destructive capability as before.
So we can say that when it comes to cannons and their balls and
smashing things to bits and terrorizing your opponents, if making
a bigger mess is better, then more mass is better, but more speed
is better-squared.
Answer:
Explanation:
When force applied to the end of the rod in perpendicular direction then net torque on the rod is given as
Now another force is applied at mid point of the rod at an angle of 30 degree with the rod
so new value of torque is given as
so we have