A slender rod is 80.0 cm long and has mass 0.120 kg. A small 0.0200-kg...

Charles Kingsley

Charles Kingsley

Answered

2022-01-04

A slender rod is 80.0 cm long and has mass 0.120 kg. A small 0.0200-kg sphere is welded to one end of the rod, and a small 0.0500-kg sphere is welded to the other end. The rod, pivoting about a stationary, frictionless axis at its center, is held horizontal and released from rest. What is the linear speed of the 0.0500-kg sphere as it passes through its lowest point?

Answer & Explanation

Janet Young

Janet Young

Expert

2022-01-05Added 32 answers

Step 1 
U=m2gy2+m3gy3=g×(0.05×(0.4)+0.02×0.4)=0.118N.m 
Determine the system's overall potential change. m2 and m3 are the masses at the rod ends. note the rod centre of mass neither gains nor loses potential. 
Step 2 
I1=2ρ0rr2dr=2×0.120.800.4r2dr=0.0064kg.m2 
Determine the rod's moment of inertia.
Step 3 
I2+3=mr2=(0.05+0.02)×0.42=0.0112kg.m2 
determine the end masses' moment of inertia
Step 4 
U=K=12(I1+I2+3)ω2 
compare the system's kinetic energy to the change in potential energy.
Step 5 
=0.118=12×0.0176ω2 
Calculate velocity at lowest point:
Step 6 
ω=3.66rads;v=rω=0.4×3.66=1.46ms

Debbie Moore

Debbie Moore

Expert

2022-01-06Added 43 answers

Answer:
Explanation:
Given that,
Slender rod
Length of rod =80cm=0.8m
Mass of slender rod =0.12kg
Sphere Bob at one end
Mass M1=0.02kg
Sphere Bod at the other end
Mass M2=0.05kg
Linear speed of mass 2 at the lowest point
We need to calculate the change in potential of the complete system. m2 and m3 are the masses at the rod ends. note the rod centre of mass neither gains nor loses potential.
So, at the lowest point,
U=M2gy2+M1gy1
Note, at the lowest point, the mass 1 is 40cm (0.4m) form the midpoint, Also, the mass 2 is 40cm(04m) from the midpoint
U=M2gy2+M1gy1
U=0.059.81(0.4)+0.029.820.4
U=0.1962+0.07848
U=0.11772Nm
Now, the moment of inertia of the rod is given as
I=r2dm
dm=2pdr
I=2pr2dr
I=2×0.120.8r2dr; from r=0  0.4
I=0.3[r33]om r=0  0.4
I=0.3[0.4330],from r=0  0.4
I=0.3×0.02133
I=0.0064kgm2.
calculating of inertia of the end masses.
I(1+2)=mr2=(m1+m2)r2
I(1+2)=(0.02+0.05)0.42
I(1+2)=0.07×0.42
I(1+2)=0.0112kgm2
Now, the Energy of the masses due to angular velocity is given as
K.E=12(I+I(1+2))w2
K.E=12(0.0064+0.0112)w2
K.E=0.0088w2
Using conservation of energy
The potential energy is equal to the kinetic energy of the system
K.E=P.E
0.0088w2=0.11772
Then, w2

karton

karton

Expert

2022-01-11Added 439 answers

Step 1
The PE of an object is the amount of energy the object will store because it is at a certain height.
The PE is given as
U=mgh
where m= mass of the object, g= acceleration due to gravity, and h= height.
Step 2
The PE of the system is given by
U=m1gh1+m2gh2=g(m1h1+m2h2)
=9.8m/s2(0.05kg×(0.4m)+0.02kg×0.4m)=0.1176Nm
Again, PE of the system is given by
U=KE=12(I3+I12)ω2ω=2UI3+I12=2U(2p00.4r2dr)+((m1+m2)r2)=2U(2×0.12kg0.8m[r33]00.4)+((0.05kg+0.02kg)(0.4m)2)=2×(0.1176Nm)(2×0.12kg0.8m[r33]00.4)+((0.05kg+0.02kg)(0.4m)2)=2×(0.1176Nm)(0.0064kgm2)+(0.0112kgm2)=3.655rad/s
The linear velocity is given by
v=rω=0.4m×3.655rad/=1.462m/s
Step 3
Hence, the required velocity is 1.462m/s

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