A regulation basketball has?

Question

a 17 cm diameter & may be approximated as a thin spherical shell. How long will it take for a basketball starting from rest to roll without slipping 3.1 m down an incline with an angle of 74.7 degrees with the horizontal (in units of s)?

Answer 1

Start by drawing your free-body diagram. The forces present are friction, normal force and weight. The weight (mg) point straight down, the normal force is perpendicular to your ramp at the point of contact with the ball (point A), and the friction is parallel to the ramp in the opposite direction of the motion (i.e. it points up the ramp) at point A.

We have no idea about the friction involved in this problem so we can eliminate the friction by summing the moments about point A. In order to do this you must break the "mg" force into its components that are parallel to the ramp and perpendicular to the ramp:
mg= mg sin(74.7) + mg cos(74.7)

Now, the normal force, the mg cos (74.7) and the friction all pass through the point A so they cause no moment. Therefore the moment about point A:

M= mg sin(74.7) r

The ball moves after we let go of it, so the moment can't be zero or it wouldn't move. By a stroke of genius, we pull the formula M=I@ (where @ is the angular acceleration) out of our butts... I mean... out of our brilliant minds. Now the formula I@ is the moment about the center of gravity, and we want the moment about point A. Using the parallel-axis theorem, the moment about point A is:

(I + mr^2) @

Therefore:
mg sin (74.7) r = (I + mr^2)@

The moment of inertia (I) for a thin-walled sphere is 2/3mr^2:
mg sin (74.7) r = (2/3mr^2 + mr^2)@

Factor out the masses:
g sin (74.7) r = (2/3r^2 + r^2)@

Congrats, you just proved that the speed of a moving object is independent of its mass... anyways, back to the problem.

@ = g sin (74.7) r / (2/3r^2 + r^2)
@ = g sin (74.7) / (2/3 + 1)r
@ = g sin (74.7) /5/3r
@ = 3g sin (74.7) /5r


The acceleration of the center of gravity as it rolls down the ramp is:

a=@r
a=@r = 3g sin (74.7) r/5r

a = 3g sin (74.7)/5

Finally, we know that the acceleration is constant (no outside forces applied to the object) and the ball started from rest (v inital is zero):

s = s initial + v initial *t + 1/2at^2
s = 0+0+ 1/2 * (3g sin (74.7)/5) *t^2
where s is the position of the object (distance)

3.1 = 3 * 9.81 * sin(74.7) /5 *t^2

I'll let you do the calculator punching... don't forget to make sure your calculator is in degree mode. I assume that the 3.1 m is the length of the incline, not the x or y distance of the ramp. If it is, you need to do the trig to find the length of the ramp.

Answer 2

hmm timing guidelines. 40 8 minutes in a recreation 12 minutes a quarter 24 sec shot clock 8 sec to get previous 1/2-courtroom 5 sec to throw ball in bounds a million min or 20 sec timeouts 6 entire timeouts plus a million 20 sec timeout it really is all i visit imagine of