A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and direction of acceleration of the stone?
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Hey mate,
# Answer- 9.91 m/s^2
# Given-
r = 80 cm = 0.8 m
f = 14 rev/25 s = 0.56 rps
# Solution-
First we'll calculate angular frequency,
ω = 2πf = 2×3.14×0.56
ω = 3.52 rad/s
Centripetal acceleration is given by formula,
a = rω^2
a = 0.8×(3.52)^2
a = 9.91 m/s^2
Centripetal acceleration exhibited by particle will be 9.91 m/s^2 along string towards centre.
Hope that helps you...
# Answer- 9.91 m/s^2
# Given-
r = 80 cm = 0.8 m
f = 14 rev/25 s = 0.56 rps
# Solution-
First we'll calculate angular frequency,
ω = 2πf = 2×3.14×0.56
ω = 3.52 rad/s
Centripetal acceleration is given by formula,
a = rω^2
a = 0.8×(3.52)^2
a = 9.91 m/s^2
Centripetal acceleration exhibited by particle will be 9.91 m/s^2 along string towards centre.
Hope that helps you...
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