A bicycle rim has a diameter of 0.65 m and a moment of inertia, measured about its center, of 0.21 kg⋅m2What is the mass of the
1 answer:
Answer:
m = 1.99 kg = 2 kg
Explanation:
The moment of inertia of a bicycle rim about it's center is given by the following formula:
where,
I = Moment of Inertia of the Bicycle Rim = 0.21 kg.m²
r = Radius of the Bicycle Rim = Diameter of the Bicycle Rim/2
r = 0.65 m/2 = 0.325 m
m = Mass of the Bicycle Rim = ?
Therefore,
<u>m = 1.99 kg = 2 kg</u>
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Answer:
q = 8.57 10⁻⁵ mC
Explanation:
For this exercise let's use Newton's second law
F = ma
where force is magnetic force
F = q v x B
the bold are vectors, if we write the module of this expression we have
F = qv B sin θ
as the particle moves perpendicular to the field, the angle is θ= 90º
F = q vB
the acceleration of the particle is centripetal
a = v² / r
we substitute
qvB = m v² / r
qBr = m v
q =
The exercise indicates the time it takes in the route that is carried out with constant speed, therefore we can use
v = d / t
the distance is ¼ of the circle,
d =
d =
we substitute
v =
r =
let's calculate
r = 2 2.2 10-3 88 /πpi
r = 123.25 m
let's substitute the values
q = 7.2 10-8 88 / 0.6 123.25
q = 8.57 10⁻⁸ C
Let's reduce to mC
q = 8.57 10⁻⁸ C (10³ mC / 1C)
q = 8.57 10⁻⁵ mC