Q6. The 600 N force applied to the bracket at As to be replaced by the two forces, Fa in the a-a direction and Fo in the b-b dir
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It is calculated that a)The angular velocity of the wheel is 272.13 rad/s,
b)On the edge of the grinding wheel, the linear speed is 47.62 m/s,
and c) On the edge of the grinding wheel, the acceleration is 12958.08 m/s².
Calculation of angular velocity, linear speed & acceleration:
Provided that,
the diameter of the wheel = 0.35 m
So, the radius, r = 0.35/2 = 0.175 m
As 1 revolution = 2π rad
(a)the angular velocity, ω = 2600 rpm = rad/s
⇒ω = 272.13 rad/s
So, the angular velocity is 272.13 rad/s.
(b)The linear speed, v = r * ω
⇒v = 0.175 * 272.13
⇒v= 47.62 m/s
(c)The angular acceleration,
⇒ = 12958.08 m/s²
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The fundamental frequency on a vibrating string is given by:
where
L is the length of the string
T is the tension
is the mass per unit length of the string
Keeping this equation in mind, we can now answer the various parts of the question:
(a) The fundamental frequency will halve
In this case, the length of the string is doubled:
L' = 2L
Substituting into the expression of the fundamental frequency, we find the new frequency:
So, the fundamental frequency will halve.
(b) the fundamental frequency will decrease by a factor
In this case, the mass per unit length is doubled:
Substituting into the expression of the fundamental frequency, we find the new frequency:
So, the fundamental frequency will decrease by a factor .
(c) the fundamental frequency will increase by a factor
In this case, the tension is doubled:
Substituting into the expression of the fundamental frequency, we find the new frequency:
So, the fundamental frequency will increase by a factor .