A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1 m, is whirled in a vertical circle with an angular velocity of 2 rero./s at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm2. Calculate the elongation of the zvire when the mass is at the lowest point of its path. Ysteel = 2 x 1011 Nm-2.

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Firstly, let's establish the key elements of the problem:

Given:

• Mass of the object (m) = 14.5 kg
• Unstretched length of the steel wire (L) = 1 m
• Angular velocity at the bottom of the circle (ω) = 2 rad/s
• Cross-sectional area of the wire (A) = 0.065 cm² = 0.065 × 10^(-4) m²
• Young's modulus of steel (Y) = 2 × 10^(11) N/m²

We are asked to find the elongation of the wire when the mass is at the lowest point of its path.

The tension in the wire at the lowest point of the circle provides the centripetal force necessary to keep the mass moving in a circular path. The tension in the wire can be calculated using the centripetal force formula:

T=m⋅v2rT=rm⋅v2

where:

• T is the tension in the wire
• m is the mass of the object
• v is the linear velocity of the object
• r is the radius of the circular path

At the lowest point of the circle, the tension in the wire must counteract both the gravitational force (mg) and provide the centripetal force (mv^2/r). So, we have:

T=mg+m⋅v2rT=mg+rm⋅v2

Given that v=r⋅ωv=r⋅ω (linear velocity = radius × angular velocity), we can rewrite the equation for tension as:

T=mg+m⋅(r⋅ω)2rT=mg+rm⋅(r⋅ω)2

T=mg+m⋅r⋅ω2T=mg+m⋅r⋅ω2

Now, we know that the elongation (ΔL) of the wire is directly proportional to the force applied and inversely proportional to the cross-sectional area and Young's modulus of the material. So, we can use Hooke's law:

F=k⋅ΔLF=k⋅ΔL

where:

• F is the force applied (in this case, the tension in the wire)
• k is the spring constant (in this case, Y⋅ALLY⋅A, where Y is Young's modulus, A is the cross-sectional area, and L is the unstretched length)
• ΔL is the elongation of the wire

Rearranging the equation, we get:

ΔL=FkΔL=kF

Substituting the values we have, we get:

ΔL=TY⋅ALΔL=LY⋅AT

ΔL=T⋅LY⋅AΔL=Y⋅AT⋅L

Now, let's substitute the expression for tension we derived earlier:

ΔL=(mg+m⋅r⋅ω2)⋅LY⋅AΔL=Y⋅A(mg+m⋅r⋅ω2)⋅L

ΔL=m⋅g⋅L+m⋅r⋅ω2⋅LY⋅AΔL=Y⋅Am⋅g⋅L+m⋅r⋅ω2⋅L

Now, let's plug in the given values and solve for ΔL:

ΔL=(14.5 kg⋅9.8 m/s2⋅1 m)+(14.5 kg⋅1 m⋅(1 m⋅2 rad/s)2)(2×1011 N/m2⋅0.065×10−4 m2)ΔL=(2×1011N/m2⋅0.065×10−4m2)(14.5kg⋅9.8m/s2⋅1m)+(14.5kg⋅1m⋅(1m⋅2rad/s)2)

ΔL=(14.5 kg⋅9.8 m/s2⋅1 m)+(14.5 kg⋅1 m⋅(1 m⋅2 rad/s)2)(2×1011 N/m2⋅0.065×10−4 m2)ΔL=(2×1011N/m2⋅0.065×10−4m2)(14.5kg⋅9.8m/s2⋅1m)+(14.5kg⋅1m⋅(1m⋅2rad/s)2)

ΔL≈142.1+0.065⋅14.5⋅(4)22×1011⋅0.065×10−4ΔL≈2×1011⋅0.065×10−4142.1+0.065⋅14.5⋅(4)2

ΔL≈142.1+3.770513×105ΔL≈13×105142.1+3.7705

ΔL≈145.870513×105ΔL≈13×105145.8705

ΔL≈1.1213×10−3 mΔL≈1.1213×10−3m

Therefore, the elongation of the wire when the mass is at the lowest point of its path is approximately 1.1213×10−31.1213×10−3 meters.

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