A wheel with a radius of 2 cm rotates according to the law φ = 0.05t^2.

Solution tasks:

Hopefully:

Wheel radius: 2 cm

Rotation law: f = 0.05t^2

Linear speed of a point on the wheel rim: 0.3 m/s

Find:

Normal and tangential acceleration of a point on the wheel rim at a given time.

Answer:

Let's convert the radius of the wheel into meters: r = 0.02 m

Let's find the moment of time t when the linear speed of a point on the wheel rim is 0.3 m/s:

0.3 m/s = r * f'(t) f'(t) = 0.3 m/s / r = 15 s^-1

Let's find the acceleration of a point on the wheel rim at a given moment in time:

f''(t) = 0.1 m/s^2 a_t = r * f''(t) = 0.002 m/s^2

The normal acceleration of a point on the wheel rim at any time is:

a_n = r * f(t)^2 = 0.02 m/s^2

Answer:

The normal acceleration of a point on the wheel rim at a given time is 0.02 m/s^2, the tangential acceleration of a point on the wheel rim at a given time is 0.002 m/s^2.

Product description

Product name: Wheel with a radius of 2 cm, rotating according to the law φ = 0.05t^2.

Description:

This digital product is a physics problem in which it is necessary to find the normal and tangential acceleration of a point on the rim of a wheel with a radius of 2 cm, rotating according to the law φ = 0.05t^2. The solution to the problem is presented in html format and presented in a readable form.

This product can be useful for students studying physics, as well as for anyone interested in mechanics and the movement of bodies.

Price: free.

This product is a solution to a physics problem involving a wheel with a radius of 2 cm, rotating according to the law φ = 0.05t^2. The problem requires finding the normal and tangential acceleration of a point lying on the rim of a wheel at the moment when its linear speed is 0.3 m/s. The solution to the problem is presented in html format and presented in a readable form.

The product description contains the conditions of the problem, formulas and laws used in the solution, calculation formulas and the answer. This product can be useful for students studying physics, as well as for anyone interested in mechanics and the movement of bodies.

The price of this product is free. If you have questions about the solution or need additional help, you can ask for help.

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A wheel with a radius of 2 cm rotates according to the law f = 0.05t^2, where f is the angular displacement in radians, t is time in seconds. Let's find the angular speed of the wheel at the moment of time when its linear speed is 0.3 m/s.

To do this, we use the formula for the relationship between linear and angular speed:

v = rω,

where v is the linear speed, r is the radius of the wheel, ω is the angular speed.

Substituting the values, we get:

0.3 m/s = 0.02 m × ω,

where

ω = 15 rad/s.

Let's find the angular acceleration of the wheel:

φ = 0.05t^2,

ω = dφ/dt = 0,1t,

α = dω/dt = 0.1 rad/s^2.

Since a point lying on the wheel rim moves in a circle, its acceleration consists of tangential and normal components:

a = at + an,

where at is the tangential acceleration directed tangentially to the circle, an is the normal acceleration directed towards the center of the circle.

Tangential acceleration can be found as the product of the wheel radius and angular acceleration:

at = rα = 0.02 m × 0.1 rad/s^2 = 0.002 m/s^2.

Normal acceleration can be found as the product of the square of the linear velocity and the radius of the wheel:

the = v^2/r = (0,3 м/с)^2/0,02 м = 4,5 м/с^2.

Thus, at the moment of time when the linear speed of a point lying on the wheel rim is 0.3 m/s, the tangential acceleration of the point is 0.002 m/s^2, and the normal acceleration is 4.5 m/s^2.

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