We are given these following data:
ω = 100000rev/min * (2πrad/rev) * (1min/60s) =
10,500 rad/s
t = 1.7 min = 102 s
r = 0.1055 m
(a) ω = ωo + at
10,500 rad/s = 0rad/s + α*102s
α = 102.94 rad/s²
(b) a_t = α r = 102.94 rad/s² * 0.1055 m = 10.86
m/s²
(c) a_r = ω² r = (10500rad/s)² * 0.1055 m = 1.16 x
10^7 m/s²
(a) Its angular acceleration is about 103 rad/s²
(b) The tangential acceleration of the point is 10.8 m/s²
(c) The radial acceleration of the point is 1.16 × 10⁷ m/s²
(d) The radial acceleration as a multiple of g is (1.18 × 10⁶) g
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Centripetal Acceleration of circular motion could be calculated using following formula:
[tex]\large {\boxed {a_c = v^2 / R} }[/tex]
a_c = centripetal acceleration ( m/s² )
v = velocity ( m/s )
R = radius of circle ( m )
Let us now tackle the problem!
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Given:
initial angular speed = ωo = 0 rad/s
final angular speed = ω = 100 000 rpm = 3333¹/₃π rad/s
elapsed time = t = 1.70 min = 102 s
radius = R = 10.55 cm = 0.1055 m
Asked:
(a) angular acceleration = α = ?
(b) tangential acceleration = a = ?
(b) radial acceleration = a_c = ?
(c) ratio of radial acceleration to gravitational acceleration = a_c : g = ?
Solution:
[tex]\alpha = ( \omega - \omega_o ) \div t[/tex]
[tex]\alpha = ( 3333\frac{1}{3} \pi - 0 ) \div 102[/tex]
[tex]\alpha = 32\frac{104}{153}\pi \texttt{ rad/s}^2[/tex]
[tex]\boxed{\alpha \approx 103 \texttt{ rad/s}^2}[/tex]
[tex]\texttt{ }[/tex]
[tex]a = \alpha R[/tex]
[tex]a = 32\frac{104}{153}\pi \times 0.1055[/tex]
[tex]\boxed{a \approx 10.8 \texttt{ m/s}^2}[/tex]
[tex]\texttt{ }[/tex]
[tex]a_c = \omega^2 R[/tex]
[tex]a_c = (3333\frac{1}{3} \pi)^2 \times 0.1055[/tex]
[tex]\boxed{a_c \approx 1.16 \times 10^7 \texttt{ m/s}^2}[/tex]
[tex]\texttt{ }[/tex]
[tex]a_c : g = (1.16 \times 10^7) : 9.8[/tex]
[tex]a_c : g \approx 1.18 \times 10^6[/tex]
[tex]\boxed{a_c \approx (1.18 \times 10^6) g}[/tex]
[tex]\texttt{ }[/tex]
[tex]\texttt{ }[/tex]
Grade: High School
Subject: Physics
Chapter: Circular Motion
