Asked by teal
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?
--------------------------------------------------------------------------------
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) Through what angle does the wheel turn in 18 s?
(b) What is the wheel's angular velocity at that time?
--------------------------------------------------------------------------------
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?
Answers
Answered by
Henry
a. d = 0.5at^2,
d = 0.5 * 0.35 * (18)^2 = 56.7rad.
b. V = at = 0.35 * 18 = 6.3m/s.
d = 0.5 * 0.35 * (18)^2 = 56.7rad.
b. V = at = 0.35 * 18 = 6.3m/s.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.