Collar P slides outward at a constant relative speed along rod AB, which rotates counterclockwise with a constant angular veloci
1 answer:
Answer:
a= 23.65 ft/s²
Explanation:
given
r= 14.34m
ω=3.65rad/s
Ф=Ф₀ + ωt
t = Ф - Ф₀/ω
= (98-0)×/3.65
98°= 1.71042 rad
1.7104/3.65
t= 0.47 s
r₁(not given)
assuming r₁ =20 in
r₁ = r₀ + ut(uniform motion)
u = r₁ - r₀/t
r₀ = 14.34 in= 1.195 ft
r₁ = 20 in = 1.67 ft
= (1.667 - 1.195)/0.47
0.472/0.47
u= 1.00ft/s
acceleration at collar p
a=rω²
= 1.67 × 3.65²
a = 22.25ft/s²
acceleration of collar p related to the rod = 0
coriolis acceleration = 2ωu
= 2× 3.65×1 = 7.3 ft/s²
acceleration of collar p
= 22.5j + 0 + 7.3i
√(22.5² + 7.3²)
the magnitude of the acceleration of the collar P just as it reaches B in ft/s²
a= 23.65 ft/s²
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Answer:
19.5°
Explanation:
The energy of the mass must be conserved. The energy is given by:
1)
where m is the mass, v is the velocity and h is the hight of the mass.
Let the height at the lowest point of the be h=0, the energy of the mass will be:
2)
The energy when the mass comes to a stop will be:
3)
Setting equations 2 and 3 equal and solving for height h will give:
4)
The angle ∅ of the string with the vertical with the mass at the highest point will be given by:
5)
where l is the lenght of the string.
Combining equations 4 and 5 and solving for ∅:
6)