Answer:
y= 240/901 cos 2t+ 8/901 sin 2t
Explanation:
To find mass m=weighs/g
m=8/32=0.25
To find the spring constant
Kx=mg (given that c=6 in and mg=8 lb)
K(0.5)=8 (6 in=0.5 ft)
K=16 lb/ft
We know that equation for spring mass system
my''+Cy'+Ky=F
now by putting the values
0.25 y"+0.25 y'+16 y=4 cos 20 t ----(1) (given that C=0.25 lb.s/ft)
Lets assume that at steady state the equation of y will be
y=A cos 2t+ B sin 2t
To find the constant A and B we have to compare this equation with equation 1.
Now find y' and y" (by differentiate with respect to t)
y'= -2A sin 2t+2B cos 2t
y"=-4A cos 2t-4B sin 2t
Now put the values of y" , y' and y in equation 1
0.25 (-4A cos 2t-4B sin 2t)+0.25(-2A sin 2t+2B cos 2t)+16(A cos 2t+ B sin 2t)=4 cos 20 t
So by comparing the coefficient both sides
30 A+ B=8
A-30 B=0
So we get
A=240/901 and B=8/901
So the steady state response
y= 240/901 cos 2t+ 8/901 sin 2t
Answer:
The torque on the child is now the same, τ.
Explanation:
- It can be showed that the external torque applied by a net force on a rigid body, is equal to the product of the moment of inertia of the body with respect to the axis of rotation, times the angular acceleration.
- In this case, as the movement of the child doesn't create an external torque, the torque must remain the same.
- The moment of inertia is the sum of the moment of inertia of the merry-go-round (the same that for a solid disk) plus the product of the mass of the child times the square of the distance to the center.
- When the child is standing at the edge of the merry-go-round, the moment of inertia is as follows:
- When the child moves to a position half way between the center and the edge of the merry-go-round, the moment of inertia of the child decreases, as the distance to the center is less than before, as follows:
- Since the angular acceleration increases from α to 2*α, we can write the torque expression as follows:
τ = 3/4*m*r² * (2α) = 3/2*m*r²
same result than in (2), so the torque remains the same.
Answer:
Explanation:
4μC will attract -9μC towards the centre and -5μC will repel it away from the centre. Both these forces are opposite to each other.
Force due to 4μC on -9μC towards the centre
= k x Q₁ Q₂/R² = 9 X 10⁹ X 4 X 10⁻⁶ X 9 X 10⁻⁶ / (1.2)² = 225 X 10⁻³ N/C
Force due to -5μC on -9μC away from the centre
= 9 x 10⁹ x 5 x 10⁻⁶x 9 x 10⁻⁶/( 0.8)² = 632.8 x 10⁻³ .N/C
Ner field =407.8 N/C.
Answer:
20 rad/s
Explanation:
mass, m = 12 kg
radius, r = 0.250 m
Moment of inertia of cylinder, I = 1/2 mr²
I = 0.5 x 12 x 0.250 x 0.250 = 0.375 kgm^2
Work done = Change in kinetic energy
Initial K = 0
Final K = 1/2 Iω²
W = 1/2 Iω²
ω² = 2W/ I = 2 x 75 / (0.375)
ω = 20 rad/s
Thus, the final angular velocity is 20 rad/s .
Note that
1 yd = 0.9144 m
Therefore,
The length of an American Football field is
(100 yds)*(09144 m/yd) = 91.44 m
Because the soccer field is 110 m long, its length exceeds the American Football Field by
100 - 91.44 = 8.56 m
or
(8.56/.9144) = 9.36 yd
This difference is equivalent to (8.56/91.44)*100 = 9.4%
Answer:
The Soccer Field is longer by
8.56 m, or
9.36 yd, or
9.4%
