A small block with mass m is placed inside an inverted cone...

A small block with mass $m$ is placed inside an inverted cone that is rotating about a vertical axis such that the time for one revolution of the cone is $T$ (Fig. 5.83$)$ . The walls of the cone make an angle $\beta$ with the vertical. The coefficient of static friction between the block and the cone is $\mu_{\mathrm{s}}$ . If the block is to remain at a constant height $h$ above the apex of the cone, what are the maximum and minimum values of $T ?$ figure can't copy

Key Concepts

Rotational Dynamics

Rotational dynamics involves understanding how objects behave when they are undergoing circular motion. In problems like this, the period of rotation is directly related to the centripetal force required to keep the object moving in a circular path, and this relates to the angular speed of the rotating system.

Force Equilibrium in Non-Inertial Frames

When analyzing motion in a rotating system, it is essential to consider the non-inertial (rotating) frame of reference. In such frames, fictitious forces, such as the centrifugal force, appear and must be balanced by real forces like gravity, the normal force, and friction. Setting up the equilibrium conditions correctly ensures that the object remains fixed relative to the rotating surface.

Static Friction

Static friction is the force that resists the initiation of sliding motion between two surfaces in contact. Its magnitude is determined by the coefficient of static friction and the normal force. In this scenario, static friction provides the necessary resistance against the component of the gravitational and inertial forces parallel to the surface, ensuring that the block does not slip.

Force Decomposition on Inclined Surfaces

Decomposing forces along and perpendicular to an inclined surface is essential for solving problems involving inclined planes or conical surfaces. By breaking down forces such as gravity, the inertial (centrifugal) force, and friction into components that align with the geometry of the surface, one can accurately apply equilibrium conditions to determine the system’s behavior.

Geometric Constraints in Conical Motion

The geometry of the conical surface imposes specific relationships between the height, the radius of the circular path, and the angle of inclination. Understanding these geometric constraints is crucial when linking the physical dimensions of the system with the forces involved. This geometric relationship helps translate between linear dimensions and angular dynamics, providing a necessary connection for determining the range of rotation periods that can maintain equilibrium.

Transcript

00:01 Now, so we've got a small block with mass m, and i've tried to draw this here.

00:03 You can see my mass m.

00:04 It's placed inside an inverted cone so that it's rotating around a vertical axis, such that the time for one revolution is t, which i've also represented on the image.

00:16 The walls of the cone make an angle beta with the vertical, which i've labeled.

00:20 The coefficient of static friction between the block and the cone is mu s.

00:27 If the block is to remain at a constant height, what are the minimum and maximum values of t? so i'm not, let's get started on this.

00:37 Our max t will be such t equals 2 pi r over v.

00:48 The block has a tendency to slide down, so static friction on the block was up.

00:54 So our mu will be up.

00:56 The forces are n, f, and g.

01:06 We're going to choose x and y and apply the second law.

01:09 So our x will equal n times the cosine of beta.

01:14 Y will be n times the sine of beta.

01:20 And this will be minus and plus times the sine of beta.

01:28 And plus f times the cosine of beta.

01:32 This will equal m a c equals m v squared over r and this minus mg has to equal zero.

01:46 So my max will be equal to coefficient of static friction times n and n times my cosine of b beta minus and it's working static friction times the sine of beta equal m times v squared over r minus zero.

02:16 And this will be n times the sine of beta plus this times n times the cosine of beta will equal mg.

02:40 This is at minus zero.

02:43 Okay...

Please give Ace some feedback