Answer:
8.25 rad/s
Explanation:
From the question. the condition for no slipping is represented as:
F = mg
[tex]\mu _ s N = mg[/tex]
However, the normal force N acting on the drum is represented as :
N = mrω²
Thus; we can say :
[tex]\mu_s mr \omega^2 = mg[/tex]
[tex]\omega^2 = \frac{g}{\mu_sr}[/tex]
[tex]\omega= \sqrt{\frac{g}{\mu_sr} }[/tex]
Given that :
the radius of the modeled cylinder = 32 cm = 0.32 m
the coefficient of static friction between the wallet and the surface of the drum [tex]\mu_s[/tex] = 0.45
mass (m) of the wallet = 210 grams.
Then:
[tex]\omega= \sqrt{\frac{9.81 \ m/s^2 }{0.45\ * \ 0.32 \ m} }[/tex]
ω = 8.25 rad/s
Thus, the minimum value the angular speed can have (ωmin) so that the wallet does not slip = 8.25 rad/s
The minimum value that can be possessed by the angular speed would be 8.25 rad/s. To understand, check the calculations below.
What information do we have:
Radius of the cylinder = 32cm
Angular speed = ω
coefficient of static friction µs=0.45
Wallet's mass = 210 grams
Using,
N = mrω²
We can find ω,
[tex]mg =[/tex] μmrω²
so,
ω = √g/μ[tex]_{s}r[/tex]
so,
ω could be determined by:
ω = [tex]\sqrt{\frac{9.81 m/s^2}{0.45 * 0.32} }[/tex]
= 8.25 rad/s
Thus, 8.25 rad/s is the correct answer.
Learn more about "Cylinder" here:
brainly.com/question/3692256
