Answer:
20.62N
Explanation:
According to the given data:
Moment of inertia 'I'= 0.097 kg m²
Number of rotation 'n'= 5
completes 5 rotations, Time 't'= 1.6 s
Distance = 7.1 cm=> 0.071 m
wheel slows down and stops, Time 'ts'= 1.3 s
First let calculate the angular velocity i.e
ω= (n2π)/t
ω= (5 x 2π)/1.6
ω= 19.625 rad/s
next is to calculate the the magnitude of the friction force on the disk with the help of torque formula i.e
τ= Iα
I = moment of inertia (kg∙m2)
α = angular acceleration (radians/s2)
Angular acceleration (α) can be defined as angular velocity (ω) divided by acceleration time (ts)
F.r =Iα----->(α= ω/ts)
F x 0.071 = 0.097 (19.625/1.3)
F= 20.62N
Therefore, the magnitude of the friction force on the disk is 20.62N
Answer:
The magnitude of the friction force is 20.63N
Explanation:
Given that,
Moment of inertia, I is 0.097 kg m²
Number of rotation, n is 5
completes 5 rotations, Time, t= 1.6 s
Distance is 7.1 cm = 0.071 m
Time, t = 1.3 s
Angular velocity is
[tex]\omega = \frac{n 2 \pi}{t}[/tex]
[tex]\omega = \frac{5 \times 2 \pi}{1.6} \\\\\omega = 19.63 rad/s[/tex]
The magnitude of the friction force on the disk with the help of torque formula
τ= Iα
I = moment of inertia (kg∙m2)
α = angular acceleration (radians/s2)
Angular acceleration (α) can be defined as angular velocity (ω) divided by acceleration time (ts)
F.r =Iα-----(α= ω/ts)
F x 0.071 = 0.097 (19.625/1.3)
F x 0.071 x 1.3 = 0.097 x 19.63
F= 20.63N
Thus, the magnitude of the friction force is 20.63N
