A 2.5 kg tribble is placed in a bucket and whirled in a 1.4 m radius vertical circle at a constant tangential speed. If the forc
1 answer:
Given that,
Mass of a tribble, m = 2.5 kg
Radius, r = 1.4 m
The force on the tribble from the bucket does not exceed 10 times its weight.
To find,
The maximum tangential speed.
Solution,
The force acting on the tribble is equal to the centripetal force.
F = 10mg
The formula for the centripetal force is given by :
v is maximum tangential speed
So, the maximum tangential speed is 11.7 m/s.
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Answer:
a) 20 rad/s
b) 6 m/s
Explanation:
b) Force acting on the wheel is 200 N
mass of the wheel is 100 kg
From Newton's second law of motion, F = m × a
Where F is the net force acting on the body
m is mass of the body
a is the acceleration of the body
By substituting the values we get, a = 2 m/s²
As acceleration is constant, we can use the below formula for calculating the final velocity of the object
v = u + a × t
Where v is the final velocity
u is the initial velocity
a is the acceleration
t is the time taken
u = 0 (∵ it starts from rest)
By substituting the values we get
v = 0 + 2 × 3 = 6 m/s
∴ Speed of center of mass after 3 seconds = 6 m/s
a) As the wheel rotates about z-axis, radius of gyration will be the radius of wheel
∴ Radius of the wheel = 300 mm
Torque acting on the wheel about axis of rotation = 300 mm × 200 N =
60 N·m
Torque = (Moment of inertia) × (angular acceleration)
Assuming that the mass of spokes of the wheel to be negligible,
Moment of inertia of the wheel about axis of rotation = 100 × 300² × = 9 kg·m²
Then,
60 = 9 × (angular acceleration)
∴ angular acceleration ≈ 6·67 rad/s²
As angular acceleration of the wheel is constant, we can use the below formula for calculation of final angular speed
=
+ α × t
Where
is the final angular velocity
is the initial angular velocity
α is the angular acceleration
t is the time taken
is 0 (∵ initially it starts from rest)
By substituting the values we get
= 6·67 × 3 = 20 rad/s
∴ Angular velocity of the wheel after three seconds = 20 rad/s
Given:
F_gravity = 10 N
F_tension = 25 N
Let's find the net centripetal force exterted on the ball.
Apply the formula:
From the given figure, the force acting towards the circular path will be positive, while the force which points directly away from the center is negative.
Hence, the tensional force is positive while the gravitational force is negative.
Thus, we have:
Therefore, the net centripetal force exterted on the ball is 15 N.
ANSWER:
15 N