Answer:
a) a = 0.477 m/s^2
b) u = 0.04862
Explanation:
Given:-
- The rotational speed of the turntable N = 33 rev/min
- The watermelon seed is r = 4.0 cm away from axis of rotation.
Find:-
(a) Calculate the acceleration of the seed, assuming that it does not slip. (b) What is the minimum value of the coefficient of static friction between the seed and the turntable if the seed is not to slip
Solution:-
- First determine the angular speed (w) of the turntable.
w = 2π*N / 60
w = 2π*33 / 60
w = 3.456 rad/s
- The watermelon seed undergoes a centripetal acceleration ( α ) defined by:
α = w^2 * r
α = 3.456^2 * 0.04
α = 0.477 m / s^2
- The minimum friction force (Ff) is proportional to the contact force of the seed.
- The weight (W) of the seed with mass m acts downwards. The contact force (N) can be determined from static condition of seed in vertical direction.
N - W = 0
N = W = m*g
- The friction force of the (Ff) is directed towards the center of axis of rotation, while the centripetal force acts in opposite direction. The frictional force Ff = u*N = u*m*g must be enough to match the centripetal force exerted by the turntable on the seed.
Ff = m*a
u*m*g = m*a
u = a / g
u = 0.477 / 9.81
u = 0.04862
Answer:
Part a)
Part b)
Explanation:
As we know that magnetic flux through the loop is given as
now we have
now rate of change in flux is given as
now we know that
Now plug in all data
Part b)
Now the radius of the loop after t = 1 s
Now plug in data in above equation
Answer:
The flux is calculated as φ=BAcosθ. The flux is thereforemaximum when the magnetic field vector is perpendicular to theplane of the loop. We may also deduce that the flux is zero whenthere is no component of the magnetic field that is perpendicularto the loop.
when angle is zero then flux is maximium because when angle zerocos is maximium