classical-mechanicsforcesfrictiongravitynewtonian-mechanics
If a body(m) slides down a frictionless inclined plane (M), will the net normal force between the ground and the inclined plane be (M+m)g ?
I feel it should be less than (M+m)g. This is because one of the component of the weight of the body(m) which is mgsinΘ does not need to be cancelled by the normal force. So, the normal force should be less than (M+m)g by mgsinΘ.
Actually I had a question in my book which said it should be (M+m)g. So, is it correct ? How ?
The weight of the object, $mg$, is split into components down the ramp and normal to the ramp. These components are $mg\sin{\theta}$ and $mg\cos{\theta}$ respectively. So to directly answer your question, the normal force is never equal to the weight of the object on an inclined plane (unless you count the limiting case of level ground). It is equal to the weight of the object times the cosine of the angle the inclined plane makes with horizontal.
When computing the acceleration of an object down a frictionless inclined plane, we are only interested in the component of force (weight) down the plane, namely $mg\sin{\theta}$. Since the plane is frictionless, there is no contribution whatsoever from the normal force.
See here to visualize how the weight of the object is split into components:
...the torque exerted by $N$ is zero but the torque exerted by $mg$ is non-zero. This means the ball must roll...
Actually, it means that the angular momentum about that axis must increase. That is not the same as rolling. If the axis is through the center of mass of the object then the only way for the angular momentum to increase is through rolling. However, if the axis does not pass through the center of mass then there is also angular momentum due to the linear motion. In other situations this is the difference between orbital angular momentum and spin angular momentum. So let's calculate the "orbital" angular momentum in this problem.
The torque is $m g R \sin(\theta)$ where $R$ is the radius of the ball and $\theta$ is the angle of the incline.
The magnitude of the "orbital" angular momentum is given by $R m v$ where $v$ is the linear velocity of the center of mass, so its time derivative is $R m a$ where $a$ is the linear acceleration of the center of mass.
From Newton's laws the linear acceleration is the component of gravity which is down the slope. This is $ma=mg \sin(\theta)$ so $a=g \sin(\theta)$.
Substituting the linear acceleration into the time derivative of the orbital angular momentum gives $R m g \sin(\theta)$ which is equal to the torque. This means that the increase in angular momentum due to the torque is fully accounted for by the increase in the "orbital" angular momentum and there is no left over torque for increasing the "spin" angular momentum. Therefore, the ball does not spin/roll regardless of which axis you examine.
Best Answer
Check out the solution in the image. It will be dependent on the inclination of the plane. Here, a is the horizontal acceleration of the inclined plane.