Question

A wheel 1.70 m in diameter lies in a vertical plane and rotates about its central axis with a constant angular acceleration of 3.60 rad/s2. The wheel starts at rest at t = 0, and the radius vector of a certain point P on the rim makes an angle of 57.3° with the horizontal at this time. At t = 2.00 s. What is the tangential speed, total acceleration, and angular position of point P.

Answer 1

To calculate the tangential speed, total acceleration, and angular position of point P on the wheel at t = 2.00 s, we can use the formulas for tangential speed, total acceleration, and angular position. By substituting the given values into these equations, we can find the required values.

Step-by-step explanation:

To calculate the tangential speed of a point on the wheel, we can use the formula:

Tangential Speed = Angular Velocity x Radius

In this case, the angular velocity is given by the equation:

Angular Velocity = Initial Angular Velocity + (Angular Acceleration x Time)

Substituting the given values and solving the equations, we can find the tangential speed, which is the speed of the point P on the rim at time t = 2.00 s.

To find the total acceleration, we can use the formula:

Total Acceleration = Tangential Acceleration + Radial Acceleration

The tangential acceleration can be calculated using the equation:

Tangential Acceleration = Angular Acceleration x Radius

The radial acceleration can be calculated using the equation:

Radial Acceleration = (Angular Velocity x Angular Velocity) x Radius

By substituting the given values into these equations, we can find the total acceleration at time t = 2.00 s.

To find the angular position of point P at time t = 2.00 s, we can use the equation:

Angular Position = Initial Angular Position + (Initial Angular Velocity x Time) + (0.5 x Angular Acceleration x Time x Time)

Substituting the given values, we can find the angular position of point P at t = 2.00 s.

Answer 2

Total acceleration will be
`30.70m/sec^2`

Step-by-step explanation:

We have given initial tangential acceleration
`\alpha =3.60rad/sec^2`

Radius r = 1.70 m

Initial angular velocity
`\omega =0rad/sec`

Time t = 2 sec

From first equation of motion

`\omega _f=\omega +\alpha t`

`\omega _f=0 +3.6* 2=7.2rad/sec`

Radial acceleration is equal to
`a_r=(v^2)/(r)=(7.2^2)/(1.7)=30.49m/sec^2`

So total acceleration will be equal to

`a=√(3.6^2+30.49^2)=30.70m/sec^2`