Question

A particle is moving along a circular path of 2-m radius such that its position as a function of time is given by u = (5t 2) rad, where t is in seconds. Determine the magnitude of the particle’s acceleration

Answer 1

The magnitude of the total acceleration of the particle at t = 2.0 s, considering both tangential acceleration and centripetal acceleration, is 253 m/s².

Step-by-step explanation:

To determine the magnitude of the particle's acceleration at t = 2.0 s, we must consider both the tangential acceleration and the centripetal acceleration. The position function of the particle is given as u = (5t²) rad. We know that the tangential acceleration is found by taking the derivative of the velocity function concerning time. However, the provided information is not directly about speed but about the angular position. First, we would derive the angular velocity, ω(t), by differentiating the angular position, u, with respect to time, which gives ω(t) = du/dt = 10t rad/s. Then the tangential acceleration, at, is the derivative of angular velocity with respect to time, so at = dω/dt = 10 m/s².

Next, we calculate the centripetal acceleration, which depends on the velocity and the radius of the circular path. Since the radius (r) is given as 2 m and the angular velocity at t = 2.0 s is ω(2) = 20 rad/s, we can find the linear velocity (v) as v = r*ω. Thus, v(2) = 2m * 20 rad/s = 40 m/s. The formula for centripetal acceleration (ac) is ac = v²/r, so ac(2) = (40 m/s)² / 2 m = 800 m/s².

The total acceleration is the vector sum of tangential and centripetal acceleration, which in this case because they are perpendicular, is determined using the Pythagorean theorem: a = √(at² + ac²). Plugging the values in, a = √(10² + 800²) m/s² = √(64100) m/s² = 253 m/s². Therefore, the magnitude of the total acceleration of the particle at t = 2.0 s is 253 m/s².

Answer 2

Step-by-step explanation:

Given

radius of path
`r=2\ m`

Velocity of Particle
`\theta =5t^2 rad`

where t=time in seconds

angular velocity of particle is given by

`\omega =\frac{\mathrm{d} \theta }{\mathrm{d} t}`

`\omega =2* 5t=10\cdot t`

And angular acceleration is given by

`\alpha =\frac{\mathrm{d} \omega }{\mathrm{d} t}`

`\alpha =10 rad/s^2`

tangential acceleration is
`a_t=\alpha * r`

`a_t=10* 2=20\ m/s^2`

Centripetal acceleration
`a_c=\omega ^2* r`

`a_c=(10t)^2* 2=200t^2`

net acceleration is sum of tangential and centripetal force at any time t is given by

`a_(net)=√((a_c)^2+(a_t)^2)`

`a_(net)=√((200t)^2+(20)^2)`

`a_(net)=20√((10t)^2+1)\ m/s`