Answer :
a)
2revs in 0.08s
so in 1s thats 25revs
therefore thats 50π radians in one second
b)
well, ω=2π/T
therefore ω=50π = 157.079rads^-1
and v=rω where r is in meters;
v=0.3x157.079
v=47.123ms^-1
c)
f=1/T
f=1/period for one rotation
1 rotation = 0.08/2 = 0.04
f=1/0.04
f=25Hz
2revs in 0.08s
so in 1s thats 25revs
therefore thats 50π radians in one second
b)
well, ω=2π/T
therefore ω=50π = 157.079rads^-1
and v=rω where r is in meters;
v=0.3x157.079
v=47.123ms^-1
c)
f=1/T
f=1/period for one rotation
1 rotation = 0.08/2 = 0.04
f=1/0.04
f=25Hz
(a) The angle of the rotation of the wheel in 1.0 seconds is 157.1 radians.
(b) The linear speed of a point on the wheel's rim is 47.13 m/s.
(c) The wheel's frequency of rotation is 1500 rpm.
The given parameters;
- radius of the wheel, r = 30 cm = 0.3 m
- angular speed of the wheel, ω = 2 rev for 0.08 s
The angle of the rotation in 1.0 seconds is calculated as;
[tex]\theta = \omega t\\\\\theta = (\frac{2 \ rev}{0.08 \ s} \times \frac{2\pi \ rad }{1 \ rev} ) \times 1\\\\\theta = 157.1 \ rad[/tex]
The linear speed of the wheel is calculated as follows;
[tex]v = \omega r\\\\v = (\frac{2 \ rev}{0.08 \ s} \times \frac{2\pi \ rad}{1 \ rev} ) \times 0.3\\\\v = 47.13 \ m/s[/tex]
The wheel's frequency of rotation is calculated as follows;
[tex]N = \frac{2 \ rev}{0.08 \ s} \times \frac{60 \ s}{1 \min} \\\\N = 1500 \ RPM[/tex]
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