Tangential acceleration of a point on the rim of the flywheel during this spin-up process is 0.2548 m/s².
Tangential acceleration is defined as the rate of change of tangential velocity of the matter in the circular path.
Given,
Radius of flywheel (r) = 1.96 cm = 0.0196m
Angular acceleration (α)= 13.0 rad/s²
The tangential acceleration formula is at=rα
where, α is the angular acceleration, and r is the radius of the circle.
using the formula; at=rα = (13.0 rad/s²) (0.0196m) = 0.2548 m/s².
The tangential acceleration is 0.2548 m/s².
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Answer:
G.P.E = 5.292 Joules
Explanation:
Given the following data;
Mass = 0.6 kg
Height = 0.9 m
To find the gravitational potential energy;
Gravitational potential energy (GPE): it is an energy possessed by an object or body due to its position above the earth.
Mathematically, potential energy is given by the formula;

Where,
G.P.E represents potential energy measured in Joules.
m represents the mass of an object.
g represents acceleration due to gravity measured in meters per seconds square.
h represents the height measured in meters.
But we know that acceleration due to gravity is equal to 9.8m/s²
G.P.E = 0.6*9.8*0.9
G.P.E = 5.292 Joules
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