A conveyer belt is driven by a large (diameter = 1.0 m) wheel. The wheel starts from rest and has an angular acceleration of 0.35 rad/s 2.

(a) Through what angle does the wheel turn in 18 s?
(b) What is the wheel's angular velocity at that time?

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On the belt is a large crate. At t = 18 s:

(c) Through what distance has the crate moved since t = 0 ?
(d) What is the crate's linear speed ?

(e) What is the acceleration of the crate?

(f) What is the radial acceleration of a small lump of grease on the rim of the wheel?
(g) A 25 g lump of grease is on the rim of the driving wheel. How large must the friction force be to keep it from flying off at t = 18 s?

a.9.02
b.30
c.28.33
d.1.26
e.6.31
f.
g.

Not sure how to go about f and g.. HELP?
also do the other answers look right?

a. d = 0.5at^2,

d = 0.5 * 0.35 * (18)^2 = 56.7rad.

b. V = at = 0.35 * 18 = 6.3m/s.

To solve this problem, we can use the following equations:

(a) θ = ½ * α * t^2
(b) ω = α * t
(c) s = r * θ
(d) v = r * ω
(e) a = r * α
(f) a_r = ω^2 * r
(g) F_friction = m * a_r

Given:
diameter of the wheel (d) = 1.0 m (radius (r) = 0.5 m)
angular acceleration (α) = 0.35 rad/s^2
time (t) = 18 s
mass of the lump of grease (m) = 25 g = 0.025 kg

(a) Through what angle does the wheel turn in 18 s?
Using the equation θ = ½ * α * t^2:
θ = ½ * 0.35 * (18)^2
θ ≈ 9.02 rad

(b) What is the wheel's angular velocity at that time?
Using the equation ω = α * t:
ω = 0.35 * 18
ω = 6.3 rad/s

(c) Through what distance has the crate moved since t = 0?
Using the equation s = r * θ:
s = 0.5 * 9.02
s ≈ 4.51 m

(d) What is the crate's linear speed?
Using the equation v = r * ω:
v = 0.5 * 6.3
v = 3.15 m/s

(e) What is the acceleration of the crate?
Using the equation a = r * α:
a = 0.5 * 0.35
a = 0.175 m/s^2

(f) What is the radial acceleration of a small lump of grease on the rim of the wheel?
Using the equation a_r = ω^2 * r:
a_r = (6.3)^2 * 0.5
a_r ≈ 19.81 m/s^2

(g) How large must the friction force be to keep the lump of grease from flying off at t = 18 s?
Using the equation F_friction = m * a_r:
F_friction = 0.025 * 19.81
F_friction ≈ 0.495 N

So, the answers are:
(a) The wheel turns approximately 9.02 radians in 18 seconds.
(b) The wheel's angular velocity is 6.3 radians per second at that time.
(c) The crate has moved approximately 4.51 meters since t = 0.
(d) The crate's linear speed is 3.15 meters per second.
(e) The crate's acceleration is 0.175 meters per second squared.
(f) The radial acceleration of a small lump of grease on the rim of the wheel is approximately 19.81 meters per second squared.
(g) The friction force needed to keep the lump of grease from flying off at t = 18 seconds is approximately 0.495 Newtons.

To find the answers to questions (f) and (g), let's first go through and verify the answers for questions (a) to (e).

(a) Through what angle does the wheel turn in 18 s?

To find the angle, we can use the equation: θ = ω₀t + (1/2)αt², where θ is the angle turned, ω₀ is the initial angular velocity, α is the angular acceleration, and t is the time.

Given:
Initial angular velocity, ω₀ = 0 (since the wheel starts from rest)
Angular acceleration, α = 0.35 rad/s²
Time, t = 18 s

Using the equation, θ = (1/2)αt²:
θ = (1/2) * 0.35 * (18)²
θ = 0.35 * 162
θ = 56.7 rad

Therefore, the wheel turns approximately 56.7 radians in 18 seconds.

(b) What is the wheel's angular velocity at that time?

To find the angular velocity, we can use the equation: ω = ω₀ + αt, where ω is the final angular velocity.

Given:
Initial angular velocity, ω₀ = 0 (since the wheel starts from rest)
Angular acceleration, α = 0.35 rad/s²
Time, t = 18 s

Using the equation, ω = ω₀ + αt:
ω = 0 + 0.35 * 18
ω = 6.3 rad/s

Therefore, the wheel's angular velocity at that time is 6.3 rad/s.

(c) Through what distance has the crate moved since t = 0?

To find the distance, we can use the equation: s = v₀t + (1/2)at², where s is the distance, v₀ is the initial velocity, a is the acceleration, and t is the time.

Given:
Initial velocity, v₀ = 0 (since the crate starts from rest)
Acceleration, a = ? (not given)
Time, t = 18 s

Unfortunately, the acceleration of the crate is not given. Could you provide the value of the acceleration?

(d) What is the crate's linear speed?

To find the linear speed, we can use the equation: v = ωr, where v is the linear speed, ω is the angular velocity, and r is the radius.

Given:
Angular velocity, ω = 6.3 rad/s
Radius, r = 1.0 m

Using the equation, v = ωr:
v = 6.3 * 1.0
v = 6.3 m/s

Therefore, the crate's linear speed is 6.3 m/s.

(e) What is the acceleration of the crate?

To find the acceleration of the crate, we need more information. Could you provide the acceleration of the crate?

Now, let's move on to questions (f) and (g).

(f) What is the radial acceleration of a small lump of grease on the rim of the wheel?

The radial acceleration of a small lump of grease on the rim of the wheel is the acceleration towards the center of the wheel. This can be calculated using the equation: ar = ω²r, where ar is the radial acceleration, ω is the angular velocity, and r is the radius.

Given:
Angular velocity, ω = 6.3 rad/s
Radius, r = 1.0 m

Using the equation, ar = ω²r:
ar = (6.3)² * 1.0
ar = 39.69 m/s²

Therefore, the radial acceleration of a small lump of grease on the rim of the wheel is 39.69 m/s².

(g) A 25 g lump of grease is on the rim of the driving wheel. How large must the friction force be to keep it from flying off at t = 18 s?

To find the friction force needed to keep the grease from flying off, we need to consider the centrifugal force acting on the grease, which is equal to the product of the mass, the radial acceleration, and the radius.

Given:
Mass, m = 25 g = 0.025 kg
Radial acceleration, ar = 39.69 m/s²
Radius, r = 1.0 m

Using the equation, Fc = m * ar * r:
Fc = 0.025 * 39.69 * 1.0
Fc = 0.99225 N

Therefore, the friction force needed to keep the grease from flying off at t = 18 s is approximately 0.99225 N.

Please note that answers (c) and (e) are dependent on the given acceleration of the crate, which is not provided. Once the value of the crate's acceleration is provided, we can calculate (c) and (e) as well.