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The linear acceleration of centre of gravity of a ball which rolls down without slipping over an inclined plane with an angle of inclination of 30° will be
- a)7 ms-2
- b)7 ms-1
- c)3.5 ms-2
- d)3.5 ms-1
Correct answer is option 'C'. Can you explain this answer?
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° is given by:
a = g*sin(θ)/(1+I/mr^2)
where:
- g is the acceleration due to gravity (9.8 m/s^2)
- θ is the angle of inclination (30°)
- I is the moment of inertia of the ball about its centre of mass
- m is the mass of the ball
- r is the radius of the ball
For a solid sphere, the moment of inertia about its centre of mass is given by:
I = (2/5)*m*r^2
Substituting this into the equation for acceleration, we get:
a = g*sin(θ)/(1+(2/5))
a = 2/7 * g*sin(30°)
a = 0.26 * g
a = 2.55 m/s^2
Therefore, the linear acceleration of the centre of gravity of the ball is 2.55 m/s^2.
a = g*sin(θ)/(1+I/mr^2)
where:
- g is the acceleration due to gravity (9.8 m/s^2)
- θ is the angle of inclination (30°)
- I is the moment of inertia of the ball about its centre of mass
- m is the mass of the ball
- r is the radius of the ball
For a solid sphere, the moment of inertia about its centre of mass is given by:
I = (2/5)*m*r^2
Substituting this into the equation for acceleration, we get:
a = g*sin(θ)/(1+(2/5))
a = 2/7 * g*sin(30°)
a = 0.26 * g
a = 2.55 m/s^2
Therefore, the linear acceleration of the centre of gravity of the ball is 2.55 m/s^2.
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The linear acceleration of centre of gravity of a ball which rolls down without slipping over an inclined plane with an angle of inclination of 30° will bea)7 ms-2b)7 ms-1c)3.5 ms-2d)3.5 ms-1Correct answer is option 'C'. Can you explain this answer?
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