Answer:
(a) 0.38 m
(b) 2.78 m/s
(c) 0.11 watt
Explanation:
mass, m = 0.3 kg
spring constant, K = 160 N/m
initial compression, d = 12 cm = 01.2 m
initial speed, u = 3 m/s
(a) Let the maximum stretch is y.
Use conservation of energy
Initial potential energy + initial kinetic energy = final potential energy
0.5 x K x d² + 0.5 x m x u² = 0.5 x K x y²
160 x 0.12 x 0.12 + 0.3 x 0.12 x 0.12 = 160 x y²
2.304 + 0.00432 = 160 y²
y = 0.38 m
y = 38 cm
(b) Let v is the maximum speed.
The speed is maximum when the stretch in the spring is zero, so by use of conservation of energy
Initial potential energy + initial kinetic energy = final kinetic energy
0.5 x K x d² + 0.5 x m x u² = 0.5 x m x v²
160 x 0.12 x 0.12 + 0.3 x 0.12 x 0.12 = 0.3 x v²
2.304 + 0.00432 = 0.3 v²
v = 2.78 m/s
(c) The time period of the spring mass system is given by
T = 0.272 second
Energy dissipated per cycle = 0.03 J
Power, P = 0.03 / 0.272 = 0.11 Watt
Answer:
w=3.05 rad/s or 29.88rpm
Explanation:
k = coefficient of friction = 0.3900
R = radius of the cylinder = 2.7m
V = linear speed of rotation of the cylinder
w = angular speed = V/R or to rewrite V = w*R
N = normal force to cylinder
N=
These must be balanced (the net force on the people will be 0) so set them equal to each other.
There are 2*pi radians in 1 revolution so:
So you need about 30 RPM to keep people from falling out the bottom
Answer:
a) h=3.16 m, b) v_{cm }^ = 6.43 m / s
Explanation:
a) For this exercise we can use the conservation of mechanical energy
Starting point. Highest on the hill
Em₀ = U = mg h
final point. Lowest point
= K
Scientific energy has two parts, one of translation of center of mass (center of the sphere) and one of stationery, the sphere
K = ½ m + ½
w²
angular and linear speed are related
v = w r
w = v / r
K = ½ m v_{cm }^{2} + ½ I_{cm} v_{cm }^{2} / r²
Em_{f} = ½ v_{cm }^{2} (m + I_{cm} / r2)
as there are no friction losses, mechanical energy is conserved
Em₀ = Em_{f}
mg h = ½ v_{cm }^{2} (m + I_{cm} / r²) (1)
h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)
for the moment of inertia of a basketball we can approximate it to a spherical shell
I_{cm} = ⅔ m r²
we substitute
h = ½ v_{cm }^{2} / g (1 + ⅔ mr² / mr²)
h = ½ v_{cm }^{2}/g 5/3
h = 5/6 v_{cm }^{2} / g
let's calculate
h = 5/6 6.1 2 / 9.8
h = 3.16 m
b) this part of the exercise we solve the speed of equation 1
v_{cm }^{2} = 2m gh / (1 + I_{cm} / r²)
in this case the object is a frozen juice container, which we can simulate a solid cylinder with moment of inertia
I_{cm} = ½ m r²
we substitute
v_{cm } = √ [2gh / (1 + ½)]
v_{cm } = √(4/3 gh)
let's calculate
v_{cm } = √ (4/3 9.8 3.16)
v_{cm }^ = 6.43 m / s