PHYSICS HELP- will award best answer!?

a) What is the rotational speed of the cylinder about its axis?

v = rω

ω = 80 / 5

ω = 16 rad/s

b) What is the magnitude and direction of the acceleration of a point on its surface?

centripetal is the only acceleration

ac = ω²r

ac = 16²(5)

ac = 1280 cm/s²

c) At the instant a certain point on its surface is at the top of the cylinder, what is the velocity of the point?

The point touching the floor has zero velocity, the cylinder axis one radius away has velocity of 80 m/s so the topmost point at a distance of two radii must have velocity of 160 cm/s

d) Repeat if the point is at the contact with the floor.

If there is no sliding, the point in contact with the floor has zero velocity.

e) Repeat if the point is midway between the top and floor and at the forward surface of the cylinder.

The front point will have an x velocity of 80 cm/s and an identical vertical downward velocity v = rω

v = √(80² + 80²)

v = 113.137... m/s

round to the proper precision.

2. The cylinder rolls up an incline at an angle of 10 degrees.

How long does it take to slow to a stop and how far does it roll up the incline?

I will assume that the cylinder does not slip on the slope

The moment of inertia for a cylinder is

I = ½mr²

The kinetic energy will convert to gravity potential energy on the slope

mgh = ½mv² + ½Iω²

mgh = ½mv² + ½(½mr²)(v/r)²

mgh = 3mv² /4

gh = 3v² /4

h = 3v² /4g

h = 3(0.80)² /4(9.8)

h = 0.049 m

h = dsinθ

d = 0.049 / sin10

d = 0.28 m ANSWER

v² = u² + 2as

a = (v² - u²) / 2s

a = (0² - 0.80²) / 2(0.28)

a = - 1.1 m/s²

t = v/a

t = 0.80 / 1.1

t = 0.71 s ANSWER

Notice that if the cylinder were sliding without friction on the slope, the decelleration would be greater at a = gsinθ or -1.7 m/s. This means the cylinder would not climb as high up the slope and the stopping time would be shorter. The rotational inertia is significant.