Respuesta :
Answer:
a. 2.645 * 10^7 m/s^2
b. 2.645 * 10^4 N
Explanation:
Parameters given:
Velocity of rod = 2010m/s
Length of rod = 15.3cm = 0.153m
Mass of object placed at the end of the rod = 1g = 0.001kg
a. Centripetal acceleration is given as:
a = (v*v)/r
Where v = velocity
r = radius of curvature.
The radius of curvature in this case is equal to the length of the rod, since the rod makes the circular path of the motion.
Hence, centripetal acceleration at the end of the rod:
a = (2010*2010)/(0.153)
a = 26432156.86 m/s^2 = 2.64 * 10^7 m/s
b. The force needed to hold the object at the end of the rod is equal to the centripetal force at the end of the rod. Centripetal force is given as:
F = ma = (m*v*v)/r
Where a = centripetal acceleration
F = 0.001 * 2.64 * 10^7
F = 2.64 * 10^4N
The Centripetal acceleration at the end of rod is [tex]\bold {2.64 x 10^7\ m/s}[/tex] and Centripetal force required to hold the object is [tex]\bold { 2.64x 10^4\ N}[/tex].
Given here:
Velocity of rod = 2010 m/s
Length of rod = 15.3cm = 0.153 m
Mass of object placed at the end of the rod = 1g = 0.001 kg
(A). Centripetal acceleration can be calculated by
[tex]\bold {a =\dfrac { v^2}{r}}[/tex]
Where
v = velocity
r = radius of curvature.
Put the values,
[tex]\bold {a = \dfrac {(2010^2)}{(0.153)}}\\\\\bold {a = 26432156.86\ m/s^2 } \\\\\bold {a = 2.64 x 10^7\ m/s}[/tex]
(B). Centripetal force can be calculated by
[tex]\bold {F = ma = \dfrac {(m v^2)}{r}}[/tex]
Where,
a = centripetal acceleration
Thus the force
[tex]\bold {F = 0.001 \times 2.64x10^7}\\\\\bold {F = 2.64x 10^4\ N}[/tex]
Therefore, the Centripetal acceleration at the end of rod is [tex]\bold {2.64 x 10^7\ m/s}[/tex] and Centripetal force required to hold the object is [tex]\bold { 2.64x 10^4\ N}[/tex].
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