Answer :
Answer:
a) the elongation of the wire when the mass is at its lowest point on the path = 0.5 cm
b) the elongation of the wire when the mass is at its highest point on the path = 0.42 cm
Explanation:
Given that;
the angular speed [tex]\omega = 120 \ rev/min[/tex]
Then converting it to rad/s ; we have:
= [tex](120 \ rev/min )(\frac{2 \ \pi \ rad }{1 \ rev} ) (\frac{1 \ min }{60 \ s} )[/tex]
= 12.57 rad/s
The cross-sectional area of the wire A = 0.014 cm²
A = (0.014 cm²) ( [tex]\frac{10^{-4} \ m^2}{1 \ cm^3}[/tex])
A = [tex]0.014*10^{-4} \ m^2[/tex]
mass (m) = 12.0 kg
R = 0.5 m
g = 9.8 m/s²
To calculate for the mass when its at the lowest point of the path; we use the Newton's second law of motion; which is expressed as:
[tex]T - mg = ma_{rad}[/tex]
where;
[tex]a_rad = ( radical \ acceleration ) = R \omega^2[/tex]
Now; we can rewrite our equation as;
[tex]T -mg = m R \omega ^2[/tex]
[tex]T = mR \omega^2 + mg[/tex]
[tex]T = m( R \omega^2 + g)[/tex]
Replacing our given values ; we have:
[tex]T = 12.0( 0.5(12.57)^2 + 9.8)[/tex]
[tex]T = 12.0( 0.5(158.0049) + 9.8)[/tex]
[tex]T = 12.0( 79.00245 + 9.8)[/tex]
[tex]T = 12.0( 88.80245)[/tex]
T = 1065.6294 N
T ≅ 1066 N
Determining the elongation [tex]\delta l[/tex] in the wire by using the equation
Y = [tex]\frac{Tl}{AY}[/tex]
Making [tex]\delta l[/tex] the subject of the formula; we have
[tex]\delta l = \frac{Tl}{AY}[/tex]
where ;
l = length of the wire
T =Tension in the wire
A = cross - sectional area
Y = young's modulus
Then;
[tex]\delta l = \frac{(1066 N)(0.5m)}{(0.014*10^{-4}m^2)(7.0*10^{10}Pa})[/tex]
[tex]\delta l[/tex] = [tex]( 0.00544 m ) *(\frac{10 ^2 cm}{1m} )[/tex]
[tex]\delta l[/tex] = 0.5 cm
Thus, the elongation of the wire when the mass is at its lowest point on the path = 0.5 cm
b)
Using Newton's second law of motion also for the mass at its highest point of the path; we have:
[tex]T +mg = m R \omega ^2[/tex]
[tex]T = mR \omega^2- mg[/tex]
[tex]T = m( R \omega^2 - g)[/tex]
Replacing our given values ; we have:
[tex]T = 12.0( 0.5(12.57)^2 - 9.8)[/tex]
[tex]T = 12.0( 0.5(158.0049)-9.8)[/tex]
[tex]T = 12.0( 79.00245 - 9.8)[/tex]
[tex]T = 12.0( 69.20245)[/tex]
T = 830.4294 N
T = 830 N
Determining the elongation [tex]\delta l[/tex] in the wire by using the equation
Y = [tex]\frac{Tl}{AY}[/tex]
Making [tex]\delta l[/tex] the subject of the formula; we have
[tex]\delta l = \frac{Tl}{AY}[/tex]
where ;
l = length of the wire
T =Tension in the wire
A = cross - sectional area
Y = young's modulus
Then;
[tex]\delta l = \frac{(830 N)(0.5m)}{(0.014*10^{-4}m^2)(7.0*10^{10}Pa})[/tex]
[tex]\delta l[/tex] = [tex]( 0.00424 m ) *(\frac{10 ^2 cm}{1m} )[/tex]
[tex]\delta l[/tex] = 0.42 cm
Thus, the elongation of the wire when the mass is at its highest point on the path = 0.42 cm